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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
A spring of force constant `1200Nm^(-1)` is mounted on a horizontal table as shown in figure. A mass of 3.0kg is attached to the free end of the spring, pulled side ways to a distance of 2.0cm and released , what is (a) the speed of the mass when the spring is compressed by 1.0cm? Potential energy of the mass when it momentarily comed to rest ? (c) Total energy of the oscillating mass. |
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Answer» `m=3kg,k=1200Nm^(-1)`, `a=0.92m` (a) `V+?, y=1.0cm=0.01m. ` Let , `y=asinomegat` `=aomegasqrt(1-sin^(2)omegat)` or `V=aomega(1-y^(2)//a^(2))^(1//2)` `=omegasqrt(a^(2)-y^(2))=sqrt((k)/(m))sqrt((a^(2)-y^(2)))` `=sqrt((1200)/(3))[(0.02)^(2)-(0.01)^(2)]^(1/(2))` (b) Max. P.E., `U_(max)-(1)/(2)momega^(2)y^(2)_(max)` `=(1)/(2)momega^(2)a^(2)=(1)/(2)ka^(2)=(1)/(2)xx1200xx(0.02)^(2)` `=0.24J` (c) At the extreme position, where `y=a` the mass just comes to res, its K.E. becomes zero `:.` Total energy `=P.E. +K.E. =0.24+0` `=0.24J` |
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| 252. |
A body of mass `m` is released from a height h to a scale pan hung from a spring. The spring constant of the spring is `k`, the mass of the scale pan is negligible and the body does not bounce relative to the pan, then the amplitude of vibration is |
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Answer» As the pan is of negligible mass, there is no loss of kinetic energy even though the collision is inelastic. The meachanical energy of the body `m` in the fiels generated by the joing action of both the gravity force and the elastic force is conserved `i.e.,DeltaE=0`. During the motion of the body `m` from the initial to the final ( positon of maxiumu compression of the spring) position `DeltaT=0`, and therefore` DeltaU=DeltaU_(gr)+DeltaU_(sp)=0` or `-mg(h+x)+(1)/(2)kx^(2)=0` On solving the quadratic equation `:` `x=(mg)/(k)+-sqrt((m^(2)g^(2))/(k^(2))+(2mgh)/(k))` As minus sign is not acceptable `x=(mg)/(k)+sqrt((m^(2)g^(2))/(k^(2))+(2mgh)/(k))` If the body `m` were at rest on the spring, the corresponding position of `m` will be its equilibrium position and at this position the resultant force on the body `m` will be zero. Therefore the equilibrium compression `Deltax` (say ) due to the body `m` will be given by `k Deltax= mg ` or `Deltax=mg//k` Therefore separation betweent the equilibrium position and one of the extremen position i.e. the sought amplitude lt brgt `a=x-Deltax+sqrt((m^(2)g^(2))/(k^(2))+(2mgh)/(k))` The mechanical energy of oscillation which is conserved equals `E=U_(extreme)`, because at the extreme position kinetic energy becomes zero . Althought the weight of body `m` is a conservative force, it is not restoring in this problem, hence `U_(extreme)` is only concerned with the spring force. Therefore `E=U_(extreme)=(1)/(2)ka^(2)=mgh+(m^(2)g^(2))/(2k)` |
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| 253. |
A body of mass m falls from a height h on to the pan of a spring balance, figure, The masses of the pan and spring are negligible. The spring constant of the spring is k. The body gets attached to the pan and starts executing S.H.M. in the vertical direction. Find the amplitude and energy of oscillation. A. `(mg)/(k)sqrt(1+(2kh)/(mg))`B. `sqrt((k)/(mg))(1+(2kh)/(mg))`C. `sqrt((mg)/(k))(1+(2kh)/(mg))`D. `(k)/(mg)sqrt(1+(kh)/(mg))` |
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Answer» Correct Answer - A When a body of mass `m` falls from height `h` onoto the pan of spring balance, let the spring of spring balance gets compressed by a length `x`. The loss of `P.E.` of the mass `=mg(h+x)`. The gain in elastic potential energy by the spring dur to compression `=(1)/(2)kx^(2)` According to law of conservation of energy, we have `mg(h+x)=(1)/(2)kx^(2)` or `(1)/(2)kx^(2)-mgx-mgh=0` or `x^(2)-(2mgx)/(k)-(2mgh)/(k)=0` Solving this quadric equation, we have `x=((2mg)/(k)+-sqrt(((2mg)/(k))^(2)+((8mgh)/(k))))/(2)` `=(mg)/(k)+-(mg)/(k)sqrt(1+(2kh)/(mg))` In equilibrium position, the spring will be compressed by distance `(mg)/(k)`. If `r` is the amplitude of oscillation then `(mkg)/(k) +r=(mg)/(k)+-(mg)/(k)sqrt(1+(2kh)/(mg))` `:.` amplitude, `r=(mg)/(k)sqrt(1+(2kh)/(mg))` |
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| 254. |
A body of mass m falls from a height h on to the pan of a spring balance, figure, The masses of the pan and spring are negligible. The spring constant of the spring is k. The body gets attached to the pan and starts executing S.H.M. in the vertical direction. Find the amplitude and energy of oscillation. |
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Answer» When a body of mass m falls from height h on to the pan of spring balance, let the spring of spring balance get compressed by a lengthx. The loss of P.E. of the mass `=mg (h+x)`. The gain in elastic potential energy by the spring due to compression `=(1)/(2)kx^(2)` Accordin to law of conservation of energy, we have `mg(h+x)=(1)/(2)kx^(2)` or `(1)/(2)kx^(2)-mgx-mgh=0` or `x^(2)-(2mg)/(k)x-(2mgh)/(k)=0` Solving this quadratic equation, we have `x=((2mg)/(k)-sqrt(((mg)/(k))^(2)+((8mgh)/(k))))/(2)=(mg)/(k)+-(mg)/(k)sqrt(1+(2kh)/(mg))` In equilibrium position, the spring will be compressed by distance `(mg)/(k)`. If r is the amplitude of oscillation then `(mg)/(k)+r=(mg)/(k)+-(mg)/(k)sqrt(1+(2kh)/(mg))` `:.` amplitude, `r=(mg)/(k)sqrt(1+(2kh)/(mg))` Energy of oscialltion `=(1)/(2)kr^(2)=(1)/(2)k((mg)/(k))^(2)(1+(2kh)/(mg))=mgh+((mg)^(2))/(2k)` |
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| 255. |
When will the motino of a simple pendulum be simple harmonic? |
| Answer» The motion of a simple pendulum will be SHM if its angular displacement `(theta)` is very small so that `sintheta~~theta`. | |
| 256. |
Show that the motino of a particle represented by `y=sinomegat-cosomegat` is simple harmonic with a period of `2pi//omega`. |
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Answer» `y=sinomegat-cosomegat=sqrt(2)[(1)/(sqrt(2))sinomegat-(1)/(sqrt(2))cosomegat]=sqrt(2)[cos((pi)/(4))sinomegat-sin((pi)/(4))cosomegat]` It represents SHM of angular frequency `omega(=2pi//T)` Time period, `T=(2pi)/(omega)` |
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| 257. |
What physical change occurs when a source of sound moves and listener is at rest? |
| Answer» When source of sound moves, wavelength of sound waves changes. | |
| 258. |
Does the change in frequency due to Doppler effect depend on (i) distance between source and observer? (ii) the fact that source is moving towards observer or observer is moving towards the source? |
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Answer» (i) No, change in frequency due to Doppler effect has nothing to do with distance between source and listener. (ii) Yes, in case of sound waves. No, in case of light waves. |
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| 259. |
If the displacement of a moving point at any time is givenby an equation of the form `y (t) = a cos omega t + b sin omega t`, shown that the motion is simple harmonic . If `a = 3 m, b = 4m and omega = 2`: determine the period , amplitude, maximum velocity and maximum acceleration. |
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Answer» `y=acos omegat+bsin omegat` Let, `a=Rsin theta and b=Rcos theta` Then, `y=Rsintheta cos omega t +R cos theta sin omega t` `y=Rsin (omega t + theta)` Velocity, `V=(dy)/(dt)=Romega cos (omegat+theta)` Acceleration, `A=(dV)/(dt)=-omega^(2)Rsin(omegat+theta)` `=omega^(2)y` It means `A prop y` and `-ve` sign shows that A is directed towards the mean position, hence the motion is SHM. Amplitude, `R=sqrt(a^(2)+b^(2))=sqrt(3^(2)+4^(2))` `=5` units Time period, `T=(2pi)/(omega)=(2pi)/(2)=3.14s` Max. velocity, `V_(max)=Romega=5xx2=10unit` Max. acceleration `=-omega^(2)R` `=-(2)^(2)xx5=-20units` |
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| 260. |
A person riding on a merry go round emits a sound wave of a certain frequency Does a person at the centre observe the Doppler effect? |
| Answer» No, as distance of source of sound from the centre remains the same. | |
| 261. |
Will there be any Doppler whift when source and listener move in the same direction with the same velocity / |
| Answer» No. This is because relative velocity of source w.r.t. listener is zero. | |
| 262. |
Four pendulums A,B,C and D are suspended from the same elastic support as shown in figure. A and C are of the same length , while B is smaller thatn A and D is larger than A. If A is given a transverse displacement, A. D will vibrate with maximum amplitudeB. C will vibrate with maximum amplitudeC. B will vibrate with maximum amplitudeD. All the four will oscillate with equal amplitude |
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Answer» Correct Answer - B Since length of pendulums A and C is same and `T=2pisqrt(l//g)`, hence their time period is same and they will have same frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude. |
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| 263. |
A particle of mass 10g is describing SHM along a straight line with a period of 4s and amplitude 12 cm. What is the kinetic energy when it is (i) 3cm (ii) 6 cm from its equilibrium position. How do you account for the difference between the two values ? `[use pi^(2)=10]` |
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Answer» Here, `m=10g, T=4s, A=12cm.` Case (i), `y=3cm, omega=(2pi)/(T)=(2pi)/(4)=(pi)/(2)rad//s` velocity, `v=omegasqrt(A^(2)-y^(2))=(pi)/(2)sqrt(144-9)` `=(pi)/(2)sqrt(135)cm//s` `KE=(1)/(2)mv^(2)=(1)/(2)xx10xx((pi^(2))/(4)xx135)` `=168.5 erg` Case (ii), `y=6cm` `v=omegasqrt(144-36)=(pi)/(2)sqrt(108)cm//s` `KE=(1)/(2)mv^(2)=(1)/(2)xx10xx((pi^(2))/(4)xx108)=1350erg` It means the KE of particle decreases when it moves from `y=3cm ` to `6 cm.` Thi is due to increase in its potential energy. |
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| 264. |
The displacement of a particle varies with time according to the relation `y=a sin omega t +b cos omega t `.A. The motion is oscillatory but not S.H.M.B. The motion is S.H.M. with amplitude `a+b`C. The motion is S.H.M. with amplitude `a^(2)+b^(2)`.D. The motion is S.H.M. with amplitude `sqrt(a^(2)+b^(2))` |
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Answer» Correct Answer - D `Y=asinomegat +bcosomegat` …(i) Let `a=Acostheta` and `b=Asintheta` Then `a^(2)+b^(2)=A^(2)(cos^(2)theta+sin^(2))A^(2)` or `A=sqrt(a^(2)+A^(2))` From (i), `y=A costheta sinomegat+Asintheta cosomegat=Asin(omegat+theta)` It is an equation of SHM with amplitude `A=sqrt(a^(2)+b^(2))` |
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| 265. |
A test tube weighing 10g and external dismeter 2.5cm is floated vertically in water by placing 20g of mercury at its bottom. The tube is depressed in water a little and then released. Find the time of oscillation. Take `g=10ms^(-2)`. |
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Answer» Total mass of the test tube and mercury, `m=10+20=30g=.030kg` Area of cross section of the test tube is `A=pir^(2)=(22)/(7)xx((2.5)/(200))^(2)=4.91xx10^(-4)m^(2)` Density of water, `rho=10^(3) kg//sm^(3)` When the tube is floating, let the tube be depressed in water by a little distance y, and released, it will execute linear SHM with initial floating position as equilibrium position or mean position. Then spring factor, `k=(F)/(y)=(Ayrhog)/(y)=Arhog` `=(4.91xx10^(-4))xx10^(3)xx10=4.91N//m`. Here inertia factor, `m=.030kg` Time period of oscialltion, `T=2pisqrt((m)/(k))=2pisqrt((.030)/(4.91))=0.49s` |
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| 266. |
A particle of mas 10g is placed in a potential field given by` U=(500x^(2)+100)` erg `//` gm. Calculate the frequency of oscillation. |
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Answer» Potential energy of 10 gram particle is `U=10(50x^(2)+100)` erg. The force acting on the particle is given by `F=-(dU)/(dr)=-(d)/(dr)(500x^(2)+1000)` `=-1000x,` But `F=m(d^(2)x)/(dt^(2))` `:.m (d^(2)x)/(dt^(2))=-1000x or (d^(2)x)/(dt^(2))=-(1000x)/(m)` `=-(1000x)/(10)=-100x` As`(d^(2)x)/(dt^(2))=-omega^(2)x, so omega^(2)=100 or omega=sqrt(100)` Frequency of oscillation, `v=(omega)/(2pi)=(sqrt(100))/(2pi)=1.59s^(-1)` |
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| 267. |
A 0.2kg of mass hangs at the end of a spring. When 0.002kg more mass is added to the end of the spring, it stretches 7 cm more. If the 0.02 kg mass is removed what will be the period of vibration of the system? |
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Answer» `k=(F)/(x)=(0.02xx9.8)/(0.07)=2.8N//m` `T=2pisqrt((m)/(k))=2pisqrt((0.2)/(2.8))=1.68s ` |
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| 268. |
The transverse displacement of a string (clamped at its both ends ) is given by `y(x,t)=0.06sin(2pix//3)cos(120pit).` All the points on the string between two consecutive nodes vibrate withA. same frequencyB. same phaseC. same energyD. different amplitude |
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Answer» Correct Answer - A::B::C The given equation is `y(x,t)=0.06 sin ((2pix)/(3))cos (120pit)` It represents a stationary wave. Therefore, all the points between two consecutive nodes. (a) vibrate with same frequency (b) in same phase, but (d) different amplitudes. The amplitude is zero at nodes and maximum at anitnodes (betweent the nodes). |
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| 269. |
Which of the following statements are true for a stationary wave ?A. Every particle has a fixed amplitude which is different from the amplitude of its nearest particleB. All the particle cross their mean position at the same timeC. There are some particles which are always at rest.D. There is no net transfer of energy across any plane. |
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Answer» Correct Answer - A::B::C::D Statements (a), (b), (c) and (d) are true as is known from the characteristaic of stationary waves. |
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| 270. |
The equation of a simple harmonic wave is given by`y = 3 sin"(pi)/(2) (50t - x)` where `x` and `y` are in meters and `x` is in second .The ratio of maximum particle velocity to the wave velocity isA. `2pi`B. `(3)/(2)pi`C. `3pi`D. `(2)/(3)pi` |
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Answer» Correct Answer - B The given equation is `y=rsin((2pit)/(T)-(2pix)/(lambda))` `(2pi)/(T)=25pi` or `T=(2)/(25)s` and `(2pi)/(lambda)=(pi)/(2) ` or `lambda=4m` Wave velocity, `upsilon=(lambda)/(T)=(4)/(2//25)=50m//s` Particle velocity, `(dy)/(dt)==3cos(25pit-(pix)/(2))xx25pi` Max. particle velocity `=((dy)/(dt))_(max)=75pi` Now, `((dy//dt)_(max))/(upsilon)=(75pi)/(50)=(3)/(2)pi` |
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| 271. |
Can a pendulum watch give correct time iin an artifical satellite ?Explain. |
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Answer» No, in an artifical satellite, the body is in weightlessness state, where`g=0`. The time period of pendulum watch. `T=2pisqrt((l)/(g))=2pisqrt((l)/(0))=oo` It means, inside the satellite pendulum does not oscillate. Hence pendulum watch connot word in satellite. |
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| 272. |
A progressive wave of frequency 500Hz is travelling with a velocity of 360 m`//` s. How far part are two points `60^(@)` out of phase ? |
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Answer» `lambda=v//n=(360)/(500)=0.72m` As, `Deltaphi=(2pi)/(lambda)Deltax` `:. Deltax=(lambda)/(2pi)Deltaphi=(0.72)/(2pi)xx(pi)/(3)=0.12m` |
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| 273. |
State and explain superposition principle. Does it apply to electromagnetic waves? |
| Answer» Yes, it is applicable to electromagnetic waves. | |
| 274. |
A pipe of length `85cm` is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below `1250 Hz`. The velocity of sound in air is `34m//s`. |
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Answer» `(a)` When the tube is closed at one end `v=(v)/( 41)(2n+1), `where` n=0,1,2,........` `=(340)/( 4 xx 0.85)(2n+1)=100(2n+1)` Thus for `n=0,1,2,3,4,5,6,....., ` we get `n_(1)=1001 H_(z), n_(2)=300 1 H_(z), n_(3)=5001 H_(z), n_(4)=700 1 H_(z),` `n_(5)=900 1 H_(z), n_(6)=1100 1 H_(z), n_(7)=1300 1 H_(z)` Since `v` shoul be `lt v_(0)=12501 H_(z)`, we nedd not go beyond `n_(6)`. `(b)` Organ pipe opened from both ends vibrates with all harmonics of the fundamental frequency. now, the fundamental mode frequency is given as `v=v//lambda` or, `v=v//2l` Here, also, end correction has been neglected. So,the frequencies of higher modes of vibrations are given by `v=n(v//2l) ...(1)` or, `v_(1)=v//2l, v_(2)=2(v//2l), v_(3)=3(v//2l)` If may be checked by patting the values of `n` in equation `(1)` that belox `1285 Hz`, there are a total of six possible natural oscillation requencies of air column in the open pipe. |
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| 275. |
The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (velocity of sound `=340ms^(-1)`).A. 4B. 5C. 7D. 6 |
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Answer» Correct Answer - D Frequency of the fundamental note produced by closed end organ pipe, `v_(0)=(upsilon)/(4l)=(340ms^(-1))/(4xx0.85m)=100Hz` Frequencies of other notes produced by pipe `v_(n)=(2n+1)v_(0)` where`n=0,1,2,3,4,.....` `=v_(0), 3v_(0), 5v_(0),7v_(0),9v_(0),11v_(0),13v_(0)` `=100Hz, 300Hz, 500Hz, 700Hz, 900Hz, 1100Hz, 1300Hz` Thus, the number of notes below the frequency `1250Hz` will 6. |
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| 276. |
Examine the following equations and test whether interference, beats or stationary waves will arise from the superposition of (i) 1 and 2 (ii) 1 and 3 (iii) 1 and 4 (iv) 2 and 3 (v) 2 and 4 (vi) 3 and 4 ? `1. xi_(1)=acos2pi((t)/(T)+(x)/(lambda))` `2. xi_(2)=acos2pi((t)/(T)-(x)/(lambda))` `3. xi_(3)=acos2pi((t)/(T)+(x+P)/(lambda)+phi)` `4. xi_(4)=acos2pi((t)/(T+alpha)+(x)/(lambda)),alphaltT` |
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Answer» (i) 1 and 2 represent two waves of same amplitude and frequency travelling in opposite directions. Their superposition will produce stationary waves. (ii) 1 and 3 represent two waves of same frequency and amplitude with a phase difference and travelling in the same direction . their supereposition will produce interference. (iii) 1 and 4 represent two waves two waves of same amplitude travelling in same direction, but having slightly different frequencies. their superposition will produce beats. (iv) Similarly, superposition of 2 and 3 will produce stationary waves. (v) Superposition of 2 and 4 will produce interference. (vi) Superposition of 3 and 4 will give rise to beats. |
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| 277. |
Is it possible to monitor the temperature of a wire by measuring its vibrational frequency? |
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Answer» Yes, `v=(p)/(2l)sqrt((T)/(m))` As, temperature increases, length increases. Therefore, frequency of vibration v decreases. Hence, changes in temperature can be monitored. |
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| 278. |
Two cubes with masses `m_(1)` and `m_(2)` were interconnected by a weightless spring of stiffness `x` and placed on a smooth horizontal surface. Then the cubes were drawn closer to each other and released simultaneously. Find the natural oscillation frequency of the system. |
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Answer» In the `C.M. ` frame `(` which is rigidly attached with the centre of mass of the two cubres `)` the cubes oscillates. We know that the kinetic energy of two body system equals `(1)/(2) mu v_(rel)^(2)`, where `mu` is the reduced mass and `v_(rel)` is the modulus of velocity of any one body particle relative to other . From the conservation of mechanical energy of the oscialltion `:` `(1)/(2) kx^(2)+ (1)/(2) mu{ (d)/(dt)(l_(0)+x)}^(2)=` constant Here `l_(0)` is the natural length of the spring. Differenting the above equation w.r.t. time, we get `:` `(1)/(2) k 2 x dot(x)+(1)/(2) mu2 dot (x) ddot (x)=0[` becomes`(d(l_(0)+x))/(dt)=dot(x)]` Thus `ddot(x)=-(k)/(mu)x(`where` mu=(m_(1)m_(2))/(m_(1)+m_(2)))` Hence the natural frequency of oscillation `: omega_(0)=sqrt((k)/(mu))`where` mu=(m_(1)m_(2))/(m_(1)+m_(2))` |
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| 279. |
One end of a long string of linear mass dnesity `8.0xx10^(-3)kgm^(-1)` is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At `t=0` the left end (fork end) of the string `x=0` has zero transverse displacement `(y=0)` and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describest the wave on the string. |
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Answer» Here,`m=8.0xx10^(-3)kgm^(-1), v=256Hz, T-90kg=90xx9.8-882N` Amplitude of wave, `r=5.0cm -0.05m.` As, the wave propagating along the string is a transverse travelling wave, the velocity of the wave is given by `upsilon=sqrt((T)/(m))=sqrt((882)/(8.0xx10^(3)))=3.32xx10^(2)ms^(-1)` `omega =2piv=2xx(22)/(7)xx256=1.61xx10^(3)rad//s` `lambda=(upsilon)/(v)=(3.32xx10^(2))/(256)m`. Propagation constant, `k=(2pi)/(lambda)=(2xx3.142xx256)/(3.32xx10^(2))=4.84m^(-1)` As, the wave is propagating along positive x direction, the equation of the wave is `y(x,t)=rsin(omegat-kx)=0.05sin(1.61xx10^(3)t-4.84x)` Here, x, y are in metre and t is in second. |
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| 280. |
For a simple pendulum, a graph is plotted its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?A. B. C. D. |
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Answer» Correct Answer - B As `K.E.=(1)/(2)momega^(2)d^(2)` and P.E. `=(1)/(2)momega^(2)d^(2)` At `d=+-A` P.E.`=` maximum , while K.E. `=0` `:. `Option `(b)` is correct. |
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| 281. |
The string of a violing emits a note of 440Hz at its correct tension. The string is bit taut and produces 4 beats per sec with a tuning fork of freq. 440Hz. Find the frequency of the note emitted by this taut string. |
| Answer» As, `nprop sqrt(T)`, therefore, freq. of vibration of string increases. As it gives 4 best `//` sec with a fork of frequency 440, therefore, its frequency would be `440+4=444Hz`. | |
| 282. |
One end of a taut string of length `3m` along the x-axis is fixed at `x = 0`. The speed of the waves in the string is `100ms^(-1)`. The other end of the string is vibrating in the y-direction so that stationary waves are set up in the string. The possible wavelength`(s)` of these sationary waves is (are)A. `y(t)=Asin((pix)/(6))cos((50pit)/(3))`B. `y(t)=Asin((pix)/(3))cos((100pit)/(3))`C. `y(t)=Asin((5pix)/(6))cos((250pit)/(3))`D. `y(t)=Asin((5pix)/(6))cos250pit` |
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Answer» Correct Answer - A::C::D There will be a node at `x=0` and antinode at `x=3` .Here, `upsilon=(omega)/(k)=100m//s.` Hence, at `x=0,y=0` and at`x=3cm,y=+A.` The above conditions are satisfied in options (a), (c) and (d). |
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| 283. |
Two simple pendulum `A` and `B` of lengths `1.69m` and `1.44m` start swinging at the time from a location where acceleration due to gravity is `10ms^(-1)`. Answer the following question. After how much time, the two pendulums will be in phase again ?A. `5.5s`B. `10.0s`C. `12.6s`D. `15.5s` |
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Answer» Correct Answer - D `t=nT_(1)=6xx2pisqrt((1.69)/(10))=12xx(22)/(7)sqrt(169)/(1000)` `=12xx(22)/(7)xx(13sqrt(10))/(100)=12xx(22)/(7)xx(13xx3.162)/(100)` `=15.5s` |
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| 284. |
Two simple pendulum `A` and `B` of lengths `1.69m` and `1.44m` start swinging at the time from a location where acceleration due to gravity is `10ms^(-1)`. Answer the following question. How many vibratinos pendulum `A` will compelte when it will be out of phase with `B` ?A. 4B. 5C. 6D. 8 |
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Answer» Correct Answer - C Let `T_(1)` and `T_(2)` be the time period of vibrations of pendulums `A` and `B` respectively. Then, `T_(1)=2pisqrt((l_(1))/(y))` and `T_(2)=2pisqrt((l_(2))/(g))` or `(T_(1))/(T_(2))=sqrt((l_(1))/(l_(2)))=sqrt((1.69)/(1.44))=(13)/(12)` If the two pendulums go out of phase in time `t`, then in time `t`, if pendulum `A` completes `n` vibrations, the pendulum `B` will complete `(n+1//2)`vibrations. Therefore, `t=nT_(1)=(n+1//2)T_(2)` or `(T_(1))/(T_(2))=((n+1//2))/(n)=(13)/(12)` or `12n+6=13n` or `n=6` |
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| 285. |
Two simple pendulum `A` and `B` of lengths `1.69m` and `1.44m` start swinging at the time from a location where acceleration due to gravity is `10ms^(-1)`. Answer the following question. After how much time, the two pendulums will be in phase again ?A. `11.0s`B. `20.0s`C. `25.2s`D. `31.0s` |
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Answer» Correct Answer - D Let `t_(1)` be the time after which two pendulums are in same phase. It means, in time `t_(1)`, if `A`complets `n` vibrations, `B` will complte `(n+1)` vibrations. So, `t_(1)=nT_(1)=(n+1)T_(2)` or `(T_(1))/(T_(2))=(n+1)/(n)=(13)/(12)` o r `12n+12=13n` or `n=12` `t_(1)=12xx2pisqrt((1.69)/(10))=31s` |
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| 286. |
It is possible to distinguish between the transverse and longitudinal waves by studying the property of |
| Answer» We have to carry out polarisation expts on them. Whereas transverse waves are polarised, longitudinal waves waves are not polarised. | |
| 287. |
A transverse harmonic disturbance is produced in a string. The maximum transverse velocity is `3m//s` and maximum transverse acceleration is `90m//s`. If the wave velocity is `20 m//s` then find the waveform. |
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Answer» Here, `v_(max)=omegar=3ms^(-1)` ...(i) `a_(max)=omega^(2)r=90ms^(-2)` velocity of wave, `v=(omega)/(k)=20ms^(-1)` `(a_(max))/(v_(max))=(omega^(2)r)/(omegar)=(90)/(3):.omega=30rad//s` From (i), `r=(3)/(omega)=(3)/(30)=0.1m` From `v=(omega)/(k)=20,k=(omega)/(20)=(30)/(20)=1.5rad//s` `:.` The equation of waveform is `y=rsin(omegat+kx)=0.1sin(30t+1.5x)` |
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| 288. |
Two tunig forks A and B produce 6 beats `//` sec. On loading A with wax number of beats remains unaltered. However, when a little of wax is removed, the number of beats becomes 2 in 2 seconds. Calculate original frequency of A if that of B is 256. |
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Answer» Here, frequency of B `=n_(B)=256` As number of beats `//`sec `=6` `:. n_(A) =n_(B)+-6=256+-6=262 or 250`. When A is loaded with wax, its frequency must have decreased from 262 to 250. That is why number of beats `//` sec remains `256-250=6`. When a little wax is removed, frequency of A increased from 250 to 255. Therefore, number of beats `//` sec `=256-255=1 or 2` beats in 2 second. Hence original frequency of `A=262Hz. ` |
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| 289. |
Find the diffference is the apparent frequencies (i) When the source approaches a stationary observer (ii) When observer is approaching the stationary source. Take b as relative velocity in each case and `upsilon` as the velocity of the waves. |
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Answer» (i) When source is approaching a stationary observer `v_(1)=(upsilon.v)/(upsilon-upsilon_(s))=(upsilonv)/(upsilon-b)` …(i) (ii) When observer is approaching th e stationary source, `v_(2)=((upsilon+upsilon_(L))v)/(upsilon)=((upsilon+b)v)/(upsilon` ...(ii) `v_(2)-v_(2)=(upsilonv)/(upsilon-b)-((upsilon+b)v)/(upsilon)` `=(upsilon^(2)v-(upsilon^(2)-b^(2))v)/((upsilon-b)upsilon)=((upsilon^(2)-upsilon^(*2)+b^(2))v)/(upsilon(upsilon-b))` `v_(1)-v_(2)=(vb^(2))/(upsilon(upsilon-b))` |
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| 290. |
A point mass is subjected to two simultaneous sinusoidal displacements in `x - direction, x_1 (t) = A sin (omega)t and x_2 (t) = A sin ((omega t + (2 pi)/(3))`. Adding a third sinusoidal displacement `x _3 (t) = B sin (omega t + phi)` brings the mas to a complete rest. The values of (B) and (phi) are.A. `sqrt(2)A, (3pi)/(4)`B. `A, (4pi)/(3)`C. `sqrt(3)A, (5pi)/(6)`D. `A, (pi)/(3)` |
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Answer» Correct Answer - B Given, `x_(1)=Asin omegat, x_(2)=A sin (omegat+(2pi)/(3))` `:. x_(1)+x_(2)=A sin omegat + A sin (omegat+(2pi)/(3))` `=Asin omegat +A[sin omegat cos ((2pi)/(3))+cos omegat sin ((2pi)/(3))]` `=A sin omegat +A[sin omegat xx(-(1)/(2))+cosomegat xx((sqrt(3))/(2))]` `=A[(1)/(2)sin omegat +((sqrt(3))/(2))cosomegat ]` `=A[cos ((pi)/(3))sin omegat+sin ((pi)/(3))cos omegat]` `=`A `sin (omegat+(pi)/(3))` As per question, `x_(1)+x_(2)+x_(3)=0` `:. x_(3)=-(x_91)+x_(2))=-Asin(omegat+(4pi)/(3))` `=A sin [omegat +(pi)/(3)+pi]=A sin (omegat+(4pi)/(3))` or B `sin (omegat+phi)=A sin (omegat+(4pi)/(3))` `:. B=A, phi=(4pi)/(3)` |
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| 291. |
A particle is moving in SHM in a straight line. When the distance of the particle from equilibrium position has values `x_(1)` and `x_(2)` , the corresponding values of velocities are `u_(1)` and `u_(2)`. Show that time period of vibration is `T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)` |
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Answer» As, `V^(2)=omega^(2)(a^(2)-y^(2)),` so `u_(1)^(2)=omega^(2)a^(2)-omega^(2)x_(1)^(2)` …(i) And `u_(2)^(2)=omega^(2)a^(2)-omega^(2)x_(2)^(2)` …(ii) Subtracting (ii) from (i), we get, `u_(1)^(2)-u_(2)^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))` or `u_(1)^(2)-u_(2)^(2)=(4pi^(2))/(T^(2))(x_(2)^(2)-x_(1)^2)` or `T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)` |
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| 292. |
A particle moves with SHM in a straight liine. In the first second afer starting from rest, it travels a distance `x_(1)` cm and in the next second it travels a distance `x_(2)` cm in the same direction. Prove that the amplitude of oscillation is `2x_(1)^(2)//(3x_(1)-x_(2))`. |
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Answer» As the particle starts from rest, it must start from the extreme position. Hence, (i) when `t=0, x=r` where r is the required amplitude. Using the relation, `x=rcosomegat,` we have `r=x_(1)=rcosomegaxx1=rcosomega` …(i) and `r-(x_(1)+x_(2))=rcosomegaxx2=rcos2omega` or `r-x_(1)-x_(2)=r(2cos^(2)omega-1)` ...(ii) Solving (i) and (ii) , we get, `r=(2x_(1)^(2))/(3x_(1)-x_(2))` |
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| 293. |
when two displacements represented by `y_(1) = a sin(omega t)` and `y_(2) = b cos (omega t)` are superimposed the motion isA. simple harmonic with amplitude `(a)/(b)`B. simple harmonic with amplitude `sqrt(a^(2)+b^(2))`C. simple harmonic with amplitude`((a+b))/(2)`D. not a simple harmonic |
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Answer» Correct Answer - B Here, `y_(1)=a sin omegat ` and `y_(2)=b cos omega t =b sin (omegat+(pi)/(2))` `A_(1)=a, A_(2)=b, phi=pi//2`j Since the frequencies of both the displacements are same, resusltant motion will be SHM. No amplitude `A=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cosphi)`. `:. A=sqrt(a^(2)+b^(2))` |
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| 294. |
A particle is moving in SHM in a straight line. When the distance of the particle from equilibrium position has values `x_(1)` and `x_(2)` , the corresponding values of velocities are `u_(1)` and `u_(2)`. Show that time period of vibration is `T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)`A. `2pisqrt((x_(2)^(2)-x_(1)^(2))/(upsilon_(1)^(2)-upsilon_(2)^(2)))`B. `2pisqrt((upsilon_(1)^(2)+upsilon_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`C. `2pisqrt((upsilon_(1)^(2)-upsilon_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`D. `2pisqrt((x_(1)^(2)+x_(2)^(2))/(upsilon_(1)^(2)+upsilon_(2)^(2)))` |
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Answer» Correct Answer - A Foa a particle undergoing SHM `upsilon=omega sqrt(A^(2)-x^(2))` `:. upsilon_(1)=omegasqrt(A^(2)-x_(1)^(2)) and upsilon_(2)=omegasqrt(A^(2)-x_(2)^(2))` Squaring we get, `upsilon_(1)^(2)=omega^(2)(A^(2)-x_(1)^(2))` and `upsilon_(2)^(2)=omega^(2)(A^(2)-x_(2)^(2))` Subtracting we get, `upsilon_(1)^(2)-upsilon_(2)^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))` or `omega=sqrt((upsilon_(1)^(2)-upsilon_(2)^(2))/(x_(2)^(2)-x_(1)^(2))) or (2pi)/(T)=sqrt((upsilon_(1)^(2)-upsilon_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` `T=2pisqrt((x_(2)^(2)-x_(1)^(2))/(upsilon_(1)^(2)-upsilon_(2)^(2)))` |
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| 295. |
Durinig the oscillation of the bob of a simple pendulum, what is the quantity that remains constant? |
| Answer» Total energy of bob in simple pendulum remains constant at all instants. | |
| 296. |
A simple pendulum of length l and with a bob of mass m moving alon a circular are of angle `theta` in a vertical plane. A sphere of mass m is placed at the end of the circular arc. What momentum will be given to the sphere by the moving bob? |
| Answer» At the end of the circular arc, the velocity to the bob of simple pendulum is zero. Therefore, the momentum of the sphere placed at the end ot circulart arc will be zero as it will not get any momentum from the bob of simple pendulum . | |
| 297. |
A ring of thin wire with active resistance `R` and inductance `L` rotates with constant angular velocity `omega` in the external uniform magneitce field perpendicular to the rotation axis. In the process , the flux of magnetic induction of external field across the ring varies with time as `Phi=Phi_(0)cos omega t`. Demonstrate that (a) the inductive current in the ring varies with time as `I=I_(m) sin ( omegat-varphi)`, where `I_(m)=._(m) Phi//sqrt(R^(2)+ omega^(2)L^(2))` with `tan varphi=omegaL//R`, (b) the mean mechanical is defined by the formula `P=1//2 omega^(2) Phi_(0)^(2)R//(R^(2)+omega^(2)L^(2))`. |
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Answer» `(a)` We have `epsilon=-(dPhi)/( dt)=omegaPhi_(0)sin omegat=Ldot(I)+RI` Put `I=I_(m) sin ( omegat-varphi)`. Then `omegaPhi_(0)sin omegat =omegaPhi_(0){sin ( omega t-varphi)cos varphi+cos (o omega-varphi)sin varphi}` `=LI_(m) omegacos ( omegat-varphi)+RI_(m) sin (omegat-varphi)` so `RI_(M) = omega Phi_(0) cos varphi` and `LI_(m) = Phi _(0) sin varphi` or ` I_(m)=( omega Phi_(0))/( sqrt(R^(2)+omega^(2)L^(2)))` and `tan varphi=( omegaL)/( R)` `(b)` Mean mechanical power required to maintain rotation `=` energy loss per unit time `=(1)/(T)int_(0)^(T)RI^(2)dt=(1)/(2) RI_(m)^(2)=(1)/(2)(omega^(2) Phi_(0)^(2)R)/( R^(2)+omega^(2)L^(2))` |
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| 298. |
Statement-1 `:` If a pendulum clock is taken to a mountain top, it gains time. Statement-2 `:` The value of acceleration due to gravity is low at mountain top than at plen.A. Statement -1 is true, Statement-2, Statement -2 is correct explanation of statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanantion of Statement-1.C. Statement-1 is correct, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - D At mountain top, the value of acceleration of gravity `(g)` is less than that a planes. Since `t prop 1//sqrt(g)`, so the time period of vibration of pendulum clock increasees i.e. it starts taking more time to complete each vibration. Therefore the pendulum clock becomes slow. |
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| 299. |
A wooden block performs SHM on a frictionless surface with frequency, `v_(0)`. The block carries a charge `+Q` on its surface. If now a uniform electric field `vec (E)` is switched on as shwon in figure., then the SHM of the block will be A. of the same frequency and with shifted mean position.B. of the same frequency and with the same mean position.C. of changes freqency and with shifted mean position.D. of changed frequency and with the same mean position.l |
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Answer» Correct Answer - A As frequency of oscillation of wooden block, `v_(0)=(1)/(2pi)sqrt((k)/(m))`, does not depend on constat external force due to electric field, hence the frequency of oscillation will be the same. But due to constant external force, the mean position of the wooden block gets shifted. Hence, option `(a)` is correct. |
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| 300. |
A and B are two wires whose fundamental frequencies are 256 and 382 Hz respectively. How many beats in 2 seconds will be heard by the third harmonic of A and secondd harmonic of B? |
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Answer» Frequency of 3rd harmonic Of A `n_(1)=3xx256=768` Frequency of second harmonic of B `n_(2)=2xx382=764` Number of beats `//` sec `=n_(1)-n_(2)=768-764=4` Number of beats in 2 sec. `=2xx4=8` |
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