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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.
| 301. |
The frequency of oscillation of a mass m suspended by a spring is `v_(1)`. If the length of the spring is cut to on-half, the same mass oscillates with frequency `v_(2)`. Determing the value of `v_(2)//v_(1)`. |
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Answer» Let the full spring be the combination of two springs in series, each of force constant k. Then, in case (i), The effective spring constant `(K_(1))` is given by `K_(1)=(kxxk)/(k=k)=(k)/(2)` Frequency of oscillation, `v_(1)=(1)/(2pi) sqrt((K_(1))/(m))=(1)/(2pi)sqrt((k)/(2m))` ...(i) In case (ii), When the spring is cut to one-half, the effective spring constant `K_(2)=k`. Frequency of oscillation, `v_(2)=(1)/(2pi)sqrt((K_(2))/(m))=(1)/(2pi)sqrt((k)/(m))` ...(ii) `:. v_(2)//v_(1)=sqrt(2)` |
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| 302. |
When the wire of a sonometer is plucked, what is the nature of waves (i) in the string (ii) in air? |
| Answer» In the string, transverse waves are formed and in air, we get longitudinal waves. | |
| 303. |
The angular velocity and amplitude of simple pendulum are `omega` and r respectively. At a displacement x from the mean position, if its kinetice energy is T and potential energy is U, find the ration of T to U. |
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Answer» Kinetice energy, `T=(1)/(2)mU^(2)=(1)/(2)momega^(2)(r^(2)-x^(2)),` Potential energy, `U=(1)/(2)momega^(2)x^(2),` so `(T)/(U)=(r^(2)-x^(2))/(x^(2))` |
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| 304. |
Why should the amplitude be small for a simple pendulum experiment? |
| Answer» When amplitude of the vibrating pendulum is small then angular displacement of the bob used in simple pendulum is small. Here, the restoring force `F=mgsintheta=mg theta=mgx//l` where x is the displacement of the bob and l is the length of pendulum.Hence, `Fpropx`. Since F is directed towards mean position, therefore the motion of the bob of simple pendulum will be SHM if `theta` is small. | |
| 305. |
What is the percentage change in the time period, if the length of simple pendulum increases by `3%`. |
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Answer» Time period of simple pendulum of length l is `T=2pisqrt((l)/(g))` …(i) When length of simple pendulum increases by `3%` then `l_(1)=l+(3)/(100)l=((103)/(100))l` `(Deltal)/(l)=(l_(1)-1)/(l)=(3)/(100)` Taking log of (i) and differentiating it , we have `(DeltaT)/(T)=(1)/(2)(Deltal)/(l)` `:. % ` change in time period `=(DeltaT)/(T)xx100=(1)/(2)(Deltal)/(l)xx100=(1)/(2)xx3=1.5%` |
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| 306. |
A simple pendulum has time period (T_1). The point of suspension is now moved upward according to the relation `y = K t^2, (K = 1 m//s^2)` where (y) is the vertical displacement. The time period now becomes (T_2). The ratio of `(T_1^2)/(T_2^2)` is `(g = 10 m//s^2)`. |
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Answer» In first case, `T_(1)=2pisqrt(l//g)` …(i) Inecond case, displacement,` y=kt^(2)` upward velocity, `v=(dy)/(dt)=2kt` upwardd acceleration , `a=(dv)/(dt)` `=2k=2xx1=2ms^(-2)` `:.` Time period, `T_(2)=2pisqrt((l)/(g+a))` …(ii) Hence `(T_(1)^(2))/(T_(2)^(2))=(4pi^(2)l)/(g)xx((g+a))/(4pi^(2)l)=(g+a)/(g)` `=(10+2)/(10)=(6)/(5)` |
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| 307. |
A 5 kg colloar is attached to a sprinig of spring constatn 500`Nm^(-1)` . It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by oscillations (b) The maximum speed and © maximum acceleration of the collar. |
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Answer» Here, `m=5kg, k=500Nm^(-1)`, `a=10cm=0.10m` (a) `T=2pisqrt((m)/(k))=2pisqrt((5)/(500))=(2pi)/(10)s=0.628s` (b) `omega=(2pi)/(T)=(2pi)/((2pi//10))=10s^(-1)` Maximum velocity `=omegaa=10xx0.10` `=1ms^(-1)` (c) Maximum acceleration `=omega^(2)a` `=(10)^(2)xx0.1=10ms^(-2)` |
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| 308. |
Abody of mass `m` is suspended from two light springs of force constant `k_(1)` and `k_(2)(k_(1))` separately. The periods of vertical oscillations are `T_(1)` and `T_(2)` respectively. Now the same body is suspended from the same two springs which are first connected in series and then in parallel. The period of vertical oscillations are `T_(s)` and `T_(p)` respectively, thenA. `T_(p)ltT_(1)ltT_(2)ltT_(s)`B. `T_(s)^(2)=T_(1)^(2)+T_(2)^(2)`C. `sqrt(T_(s))=sqrt(T_(1))+sqrt(T_(2))`D. `(1)/(T_(p)^(2))=(1)/(T_(1)^(2))+(1)/(T_(2)^(2))` |
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Answer» Correct Answer - A::B::D `T_(1)=2pisqrt((m)/(k_(1))), T_(2)=2pisqrt((m)/(k_(2)))` `T_(s)=2pisqrt((m)/((k_(1)k_(2))//(k_(1)=k_(2))))` `T_(p)=2pisqrt((m)/((k_(1)+k_(2))))` From above we note that `T_(s)gtT_(2)gtt_(1)gtT_(p),i.e.,` option (a) is correct. `(1)/(T_(p)^(2))=(k_(1)+k_(2))/(4pi^(2)m)=(k_(1))/(4pi^(2)m)+(k_(2))/(4pi^(2)m)` `=(1)/(T_(1)^(2))+(1)/(T_(2)^(2)),i.e.,` option (d) is correct. `T_(s)^(2)=(4pi^(2)m(k_(1)+k_(2)))/(k_(1)k_(2))=(4pi^(2)m)/(k_(2))+(4pi^(2)m)/(k_(1))` `=T_(2)^(2)+T_(1)^(2), i.e.,` option (b) is correct. |
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| 309. |
Determine the period of small longitudinal oscillations of a body with mass `m` in the system shown in figure. The stiffness values of the springs are `x_(1)` and `x_(2)`. The friction and the masses of the springs are negligible. |
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Answer» The net unbalanced force on the block `m` when it is at a small horiozontal distance `x` from the equilibrium position becomes `(k_(1)+k_(2))x` From `F_(x)=mw_(x)` for the block `:` `-(k_(1)+k_(2))x=m ddot(x)` Thus` ddot (x)=-((k_(1)+k_(2))/(m))x` Hence the sought time period `T=2pi sqrt((m)/(k_(1)+k_(2)))` Alternate `:` Let us set the block `m` in motion to perform small oscillation . Let us locate the block when it is at a distance `x` from its equilibrium position. As the spring force is restoring conservative force and deformation of both the springs are same, so from the conservation of mechanical energy of the spring `-` block system `:` `(1)/(2)m ((dx)/(dt))^(2)+(1)/(2) k_(1)x^(2)+(1)/(2)k_(2)x^(2)=` Constant Differentiating with respect to time `(1)/(2) m 2 ddot(x)ddot(x)+(1)/(2)(k_(1)+k_(2))2xdot(x)=0` `ddot(x)=-((k_(1)+k_(2)))/(m)x` Hence the sought time period `T=2pisqrt((m)/(k_(1)+k_(2)))` |
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| 310. |
A system of springs with their spring constants are as shown in figure. What is the frequency of oscillations of the mass m ? |
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Answer» Here, `k_(3)` and `k_(4)` are in parallel and they together are in series with `k_(1)` and `k_(2)` . The effective spring constant of the system is `(1)/(K)=(1)/(k_(1))+(1)/(k_(2))+(1)/((k_(3)+k_(4)))` `=(k_(2)(k_(3)+k_(4))+k_(1)(k_(3)+k_(4))+k_(1)k_(2))/(k_(1)k_(2)(k_(3)+k_(4)))` or `K=(k_(1)k_(2)(k_(3)+k_(4)))/((k_(1)+k_(2))(k_(3)+k_(4))+k_(1)k_(2))` Frequency of oscillation, `v=(1)/(2pi)sqrt((K)/(m))` `=(1)/(2pi)sqrt((k_(1)k_(2)(k_(3)+k_(4)))/({(k_(1)+k_(2))(k_(3)+k_(4))+k_(1)k_(2)}m))` |
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| 311. |
Find the period of small vertical oscillations of a body with mass `m` in the system illustrated in figure. The stiffness values of the springs are `x_(1)` and `x_(2)`, their masses are negligible. |
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Answer» During the vertical oscillation let us locate the bock at a vertical down distance `x` from its equilibrium position . At his moment if `x_(1)` and `x_(2)` are the additional or further elongation of the upper `&` lower springs relative to the equilibrium position, then the net unbalanced force on the block will be `k_(2) x_(2)` directed in upward direction. Hence `-k_(2)x_(2)=m ddot(x) ............(1)` we also have `x=x_(1)+x_(2) ......(2)` As the springs are massless and initially the net force on the spring is also zero so for the spring `k_(1)x_(1)=k_(2)x_(2).......(3)` Solving the Eqns `(1), (2)` and `(3)` simultaneoulsy, we get `-(k_(1)k_(2))/(k_(1)+k_(2))x=m ddot(x)` Thus`ddot(x)=-((k_(1)k_(2)//k_(1)+k_(2)))/(m)x` Hence the sought time period `T=2pisqrt(m(k_(1)+k_(2))/(k_(1)k_(2)))` |
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| 312. |
A ball of radius r is made to oscillate in a bowl of radius R, find its time period of oscillation. |
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Answer» Here, equivalent length of simiple pendulum `=` distance between centre of ball to centre of bown, i.e., `l=(R-r)` `:.` time period of oscillation of ball `=2pisqrt((R-r)//g)` |
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| 313. |
Calculate th eperiod of small oscillations of a hydrometer which was slightly pushed down in the vertical direction. The mass of the hydrometer is `m=50g`, the radius of its tube is `r=3.2mm`, the density of the liquuid is `rho=1.00 g//cm^(3)` . The rsistance of the liquid is assumed to be negligible. |
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Answer» If the hydromoter were in equlibrium or floacting, its weight will be balanced by the buoyancey force acting on it by the fluid. During its small oscillation, let us locate the hydrometer when it is at a vertically downward distance `x` from its equilibrium position. Obviously teh net unbalanced force on the hydrometer is the excess buoyancy force directed upward and equals `pir^(2)x rho g`. Hence for the hydrometer. `m ddot(x)=-pir^(2)rho g x` or, `ddot(x) =-(pi r^(2) rho g )/(m)x` Hence the sought time period `T=2alpha sqrt((m)/(pir^(2)rho g))=2.5 s.` |
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| 314. |
Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point isA. simple harmonic motionB. non-periodic motionC. periodic motionD. periodic but not S.H.M. |
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Answer» Correct Answer - A::C A ball bearing when released a little above the lower limit inside a smooth curved ball, will execute SHM with a definite period. |
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| 315. |
When a body of mass 2.0 kg is suspended by a spring, the spring is stretched. If the body is pulled down slightly and released, it coscillated up and down. What force is applied on the body by the spring when it passes through the mean position ? `(g=9.8Nkg^(-1))` |
| Answer» Since, there is no acceleration in the body at the mean position, hence the resultatn force on it will be zero, i.e., the force applied by the spring will be exactly equal to the weight of the body. | |
| 316. |
If two wave of the same frequency differ in amplitude and are propagated in opposite directions through a medium , will they produce standing waves ? Is energy transported ? Are there any nodes ? |
| Answer» Yes, stationary waves are formed. No energy is transferred is the region of stationary waves. There are no nodes, but there are positions of minimum amplitude. | |
| 317. |
An organ pipe is in resonance with a tuning fork. What change will have to be made in the length l to maintain resonance if (i) temperature increases (ii) air is replaced by hydrogen (iii) pressure is made higher ? |
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Answer» If organ pipe is open, its fundamental frequency is `v=(upsilon)/(2l)=(1)/(2l)sqrt((lambdaP)/(rho))` (i) If temperature increase , `upsilon` increases. To keep v constant, l must be incresed. (ii) When air is replaced by hydrogen, `rho` decreases. To keep v constant, l must be increased. (iii) When pressure is made higher, `rho` increases and `P//rho` remains constant. Therefore, no change is required in resonance length. |
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| 318. |
What are the conditions for resonance of air column with a tuning fork? |
| Answer» The resonance will take place in air column if the compression or rarefaction produced by the vibration of the tuning fork travels from open end of the air column to lower closed end (water level in the resonance apparatus) and back to the upper open end in the same time in which the prong goes from one extremen to other. | |
| 319. |
The equation of a plane progressive wave is give by equation `y=10sin2pi(t-0.005x)` where y and x are in cm and t is in seconds. Calculate the amplitude, frequency, wavelength and velcoity of the wave. |
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Answer» The given equation is `y=10sin2pi(t-0.005x)` `=10sin2pixx0.005((t)/(0.005)-x)` `y=10sin2pixx(1)/(200)(200t-x)` Compare it with the standard equation of a travelling wave `y=rsin(2pi)/(lambda)(vt-x)` `r=10cm`. This is the amplitude, `lambda=200cm=2metre` `v=200 cm/s=2m//s` Frequency, `v-(upsilon)/(lambda)=(2)/(2)=1Hz` |
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| 320. |
A wave is expressed by the equation `y=0.5sinpi(0.01x-3t)` where y and x are in metre and t in seconds. Find the speed of propagation. |
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Answer» The given equation is `y-0,5sinpi(0.01x-3t)` `=0.5sin((pix)/(100)-3pit)` Compare it with the standard from of equation `y=rsin((2pix)/(lambda)-(2pit)/(T))` `r=0.5m, (2pi)/(lambda)=(pi)/(100),lambda=200m. ` `(2pi)/(T)=3pi,T=(2)/(3)sec.` Speed of propagation ` upsilon=vlambda=(lambda)/(T)` `upsilon=(200)/(2//3)=300m//s` |
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| 321. |
A sonometer wire is vibrating is resonance with a tuning frok. Keeping the tension applied same, the length of the wire is dourbled. Under what conditions would the tuning fork still be in resonance with the wire? |
| Answer» When tension is kept the same, wire of twice the length vibrates in its second harmonic. Thus if tuning fork resonates at L, it will resonate at 2L. | |
| 322. |
Calculate the fundamental frequency of a sonometer wire fo length 20cm, tension 25N, cross sectional area `10^(-2)cm^(2)` and density of material `=10^(4)kg//m^(3)`. |
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Answer» `v_(1)=?l=20cm=(1)/(5)m,T=25N,` `a=10^(2)cm^(2)=10^(-6)m^(2), rho=10^(4)kg//m^(3)` Mass per unit length of wire, `m=axx1xxrho=10^(-6)xx1,rho=10^(4)=10^(-2)kgm^(-1)` `v_(1)=(1)/(2l)sqrt((T)/(m))=(1)/(2xx1//5)sqrt((25)/(10^(-2)))=(5)/(2)xx50` `=125Hz` |
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| 323. |
A sonometer wire has a length of 114 cm between its two fixed ends. Where should the two briges be places so as todivide the wire into three segments, whos fundamental frequencies are in the ratio `1:3:4?` |
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Answer» Here, `L=114cm`. Let the two bridges divide the entire length in three parts of lengths `l_(1), l_(2), l_(3)` such that `n_(1):n_(2):n_(3)=1:3:4.` As `nprop(1)/(l), ` therefore, `l_(1):l_(2):l_(3)=(1)/(1):(1)/(3):(1)/(4)=12:4:3.` `l_(1)=(Lxx12)/((12+4+3))=(114)/(19)xx12=72cm` `l_(2)=(114xx4)/(19=24cm` , `l_(3)=(114xx3)/(19)=18cm` Hence two bridges must be placed at 72cm and `(72+24)=96cm` from one end. |
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| 324. |
The following equation represents standing wave set up in a medium , `y = 4 cos (pi x)/(3) sin 40 pi t` where `x and y` are in cm and t in second. Find out the amplitude and the velocity of the two component waves and calculate the distance adjacent nodes . What is the velocity of a medium particle at ` x = 3 cm` at time `1//8 s`? |
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Answer» From the knowledge of theory, if equation of statinary wave is `y=2Acoskxsinomegat,` then the component waves are `y_(1)=Asin(omegat-kx)` and `y_(2)=Asin(omegat+kx)` `:.` Amplitude of component waves `=A=(2A)/(2)=(4)/(2)=2cm` velocity of component waves `v=(omega)/(k)=(40pi)/(pi//3)=120cm//s=1.2m//s` (b) Wavelength, `lambda=(2pi)/(k)=(2pi)/(pi//3)=6cm.` Distance between adjacent nodes `=(lambda)/(2)=(6)/(2)=3cm` (c) Particle velocity `upsilon=(dy)/(dt)=(d)/(dt)[4cos((pix)/(3))sin40pit]` `160picos((pix)/(3))cos40pit` For `x=3cm ` and `t=(1)/(8)s` `upsilon=160picos((pixx3)/(3))cos40pixx(1)/(8)` `=160pi(-1)(-1)=160picm//s` `=1.6pim//s` |
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| 325. |
Find the time dependece of the angle of deviation of a mathematical pendulum `80 cm` in length if at the initial moment the pendulum. (a) was deviated through the angle `3.0` and then set free without push, `(b)` was in the equlibrium position and its lower end was imparted the horizontal velcoity `0.22m//s,` (c ) was deviated throught the angle `30^(@)` and its lower end was imparted the velocity `0.22m//s` derected toward the equilibrium position. |
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Answer» The natural angular frequency of a mathematical pendulum equals `omega_(0)sqrt(g//l)` `(a)` We have the solution of S.H.M. equation in angular form `: ` `theta=theta_(m)cos (omega_(0)t+alpha)` If at the initial moment `i.e.` at `t=0, theta=theta_(m)` than` alpha=0` Thus the above equation takes the form `theta=theta_(m)cos omega_(0)t` `=theta_(m)cossqrt((g)/(l))t=3^(@)cos sqrt((9.8)/(0.8))t` Thus `theta=3^(@) cos 3.5 t` `(b)` The S.H.M. equation in angular form `:` `theta=theta_(m)sin (omega_(0)t+alpha)` If at the initial moment` t=0, theta=0, ` then `alpha=0` . Then the above equation takes the form `theta=theta_(m)sin omega_(0)t` Let `v_(0)` be the velocity of the lower end of pendulum at `theta=0`, then form conserved of mechanical energy of oscillation `E_(mean)=E_(extreme)` or `T_(mean )=U_(extreme)` or, `(1)/(2) m v_(0)^(2)=mgl(1-cos theta_(m))` Thus `theta_(m)=cos^(-1)(1-(v_(0)^(2))/(2gl))=cos^(-1)[1-(0.22)^(2)/(2xx9.8xx0.8)]=4.5^(@)` Thus the sought equation becomes `theta= theta_(m) sin omega_(0)t=4.5^(@) sin 3.5 t` `(c)` Let `theta_(0)` and `v_(0)` be the angular deviation and linear velocity at `t=0`. As the mechanical energy of oscillation of the mathematical pendulum is conservation `(1)/(2)m v_(0)^(2)+mgl (1-cos theta_(0))=mgl (1-cos theta_(m))` or `(v_(0)^(2))/(2)=gl(cos theta_(0)-cos theta_(m))` Thus `theta_(m)=cos^(-1){cos theta_(0)-(v_(0)^(2))/(2 gl)}=cos ^(-1){cos 3 ^(@)-((0.22)^(2))/(2xx9.8xx0.8)}=5.4^(@)` Then from `theta=5.4^(@) sin (3.5 t+alpha)`, we see that ` sin alpha=(3)/(5.4)` and `cos alpha lt 0` because the velocity is directed towards the centre . Thus `alpha=(pi)/(2)+1.0` radians and we get the answer. |
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| 326. |
Sound travels faster on a rainy day than on a dry day. Does it imply that speed of sound in moist hydrogen is greater than in dry hydrogen ? |
| Answer» No. This is because density of water vapour is more than that of hydrogen. Therefore, density of moist hydrogen is more than density of dry hydrogen. Therefore, speed of sound in moist hydrogen is less that in dry hydrogen. | |
| 327. |
A uniform cylinder of length (L) and mass (M) having cross sectional area (A) is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half - submerged in a liquid of density (rho) at equilibrium position. When the cylinder is given a small downward push and released it starts oscillating vertically with small amplitude. If the force constant of the spring is (k), the prequency of oscillation of the cylindcer is. |
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Answer» As the cylinder is half submerged in a liquid of density` rho` so upthrust on cylinder `=A(L//2)rhog` . If l is the extension of spring in equilibrium position, then `kl=Mg-A(L//2)rhog` ..(i) If y is the displacement of cylinder from equilibrium position downwards, then restoring force `F=[k(l+y)-{Mg-A(L//2+y)rhog}]` `-[Mg-(AL)/(2)rhog+ky-Mg+A((L)/(2)+y)rhog]` `[`From (i) `]` `=-[ky+Ayrhog)=-(k+Arhog)y` ...(ii) From (ii), `Fpropy` and `-ve` sign show that the force F is directed towards the mean position, hence the cylinder if left free will execute SHM. Comparing (ii) with the relatin, `F=-Ky,` we have spring factor, `K=(k+Arhog)` Inertia factor `=` mass of cylinder `=` M Frequency of oscillation, `v=(1)/(2pi)sqrt((spri ng fact o r)/(i n e rtia fact o r))` or `v=(1)/(2pi)sqrt((k+Arhog)/(M))` |
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| 328. |
A uniform cylinder of length (L) and mass (M) having cross sectional area (A) is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half - submerged in a liquid of density (rho) at equilibrium position. When the cylinder is given a small downward push and released it starts oscillating vertically with small amplitude. If the force constant of the spring is (k), the prequency of oscillation of the cylindcer is.A. `(1)/(2pi)[(K-Adg)/(M)]^(1//2)`B. `(1)/(2pi)[(K+Adg)/(M)]^(1/2)`C. `(1)/(2pi)[(K+Adg)/(M)]^(1//2)`D. `(1)/(2pi)[(K-Adg)/(M)]^(1/2)` |
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Answer» Correct Answer - C When the cylinder is given a downward push through a distance `y`, the restoring upward force in the spring `=-Ky.` Additional upward force due to buoyancy `=-Aydg.` Net upward restoring force, `F=-(ky+Aydg)` `=-(K+Adg)y` As, `F prop y` and it is directed towards equlibrium position, hence the cylinder will execute linear S.H.M. Here, spring factor `=(KAdg)` inertia factor`=` mass of cylinder `=M` Frequency, `v=(1)/(2pi)sqrt ((spr i ng fact o r)/(i n ertia fact o r))` `=(1)/(2pi)sqrt((K+Adg)/(M))` |
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| 329. |
There are two damped oscilaltions with the following period` T` and damping coefficients `beta: T_(1)=0.10 ms, beta_(1)=100 s^(-1)` `T_(2)=10 ms, beta=10 s^(-1)`. Which of them decays faster ? |
| Answer» The first oscillation decays faster in time. But if one takes the natural time scale, the period `T` for each oscillation, the second oscillation attenuates faster during that period. | |
| 330. |
A point participates simultaneously in two harmonic oscillations of the same direction`:x_(1) =a cos omega t ` and `x_(2)=a cos 2 omega t. ` Find the maximum velocity of the point . |
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Answer» Given,` x_(1)= a cos omegat ` and `x_(2)= a cos 2 omegat ` so, the net displacement, `x=x_(1)+x_(2)=a{cos omegat +cos 2 omegat}=a{cos omegat+2 cos ^(2) omegat-1}` and `v_(x)=x=a{-omega sin omegat -4 omega cos omegat sin omega t }` For `x` to be maximum, ` ddot x=a omega^(2)cos omegat -4 a omega^(2)cos^(2)omegat+4 a omega^(2)sin ^(2)omegat=0` or, `8 cos^(2)omegat+cos omegat-4=0,` which is a quadratic equation for `cos omegat`. Solving for accepctable value ` cos omegat =0.644` thus `sin omegat=0.765` and `v_(max)=|v_(x_(max))|=+a omega[0.765=4xx0.765xx0.644]=+2.73 a omega` |
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| 331. |
Under what condition does a sudden phase reversal of waves occur on reflection? |
| Answer» On reflection from a denser medium, waves undergo a sudden phase reversal. | |
| 332. |
A perosn normally weighting 50kg stands on a massless platform which oscillates up and down harmonically at a frequency of `2.0s^(-1)` and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time. (a) Will there be any in weight of body , during the oscillation? (b) If answer to part (a) is yes, what will be the maximum and minimum readin in the machine and at which position? |
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Answer» (a) Yes, We know that when a platform on which a perosn of mass m is standing, is accelerating upwards with acceleration a, then the effective weight of the person becomes`=m(g+a)`. If the platform is accelerating downwards with acceleration a, then effective weight of persono on platform `=m(g-a)` . During oscillation of a platform, a perosn standing on it, will be accelerated upwards and downwards with time. Due to it, the weight of a person on platform changes during the oscillation. (b) Max. acceleration of platfor, `a_(max.)=omega^(2)r=4pi^(2)v^(2)r=4xx(3.14)^(2)xx2^(2)xx(5xx10^(-2))=7.89m//s^(2)` Maximum reading in machine while moving upwards `=m(g+a)=50(9.8+7.89)=884.5N` Minimum readin in machine while moving downwards `=m(g-a)50(9.8-7.89)=95.5N` |
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| 333. |
In the above experiment, the value of end correction isA. `0.8 cm`B. `0.4 cm`C. `0.6 cm`D. `0.2 cm` |
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Answer» Correct Answer - B End correction, `x=(l_(2)-3l_(1))/(2)=(47.6-3xx15.6)/(2)=(0.8)/(2)=0.4cm` |
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| 334. |
A student plots the frequency `(n)` of given length of a wire stretched by a given force versus diameter `(D)` of the wire. The correct plot figure isA. B. C. D. |
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Answer» Correct Answer - C From `n=(1)/(lD)sqrt((T)/(pirho))`, we find that `n prop (1)/(D0`, when `l, T` and `rho` are fixed. `:.` Graph between `n` and `1//D` is a straight line as shown in `(c)`. |
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| 335. |
A particle executes SHM. Then the graph of velocity as a function of displacement isA. a straight lineB. a circleC. an ellipseD. a hyperbola |
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Answer» Correct Answer - C an ellipse |
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| 336. |
If a graph is drawn for acceleration versus displacemnet of SHM. It is aA. straight lineB. circleC. ellipseD. hyperbola |
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Answer» Correct Answer - A straight line |
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| 337. |
When are the velocity and acceleration in the same direction SHM? |
| Answer» When a particle executing SHM Is moving from extreme position towards mean position, then the velocity and acceleration are in the same direction,i.e., towards the mean position. | |
| 338. |
When are the diplacement and velocity in the same direction in SHM? |
| Answer» When a particle executing SHMi is moving from mean position towards extremen position, then the displacement and velocity are in the same direction , i.e., away from the mean position. | |
| 339. |
A block is resting on a piston which is moving vertically with simple harmonic motion of period 1.0 second. At what amplitude of motion will the block and piston separate? What is the maximum velocity of the piston at this amplitude? |
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Answer» The block and piston will just separate if `a_(max)=g or omega^(2)A=g or (4pi^(2))/(T^(2))A=g` or `A=(T^(2)g)/(4pi^(2))=((1.0)^(2)xx9.8)/(4xx(22//7)^(2))=0.248m` Max. velocity of the block is `v_(max)=omega=(2pi)/(T)A=(2xx3.142)/(1.0)xx0.248` `=1.56ms^(-1)` |
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| 340. |
A wave `y = a sin (omegat - kx)` on a string meets with another wave producing a node at `x = 0`. Then the equation of the unknown wave isA. `y=asin(omegat+kx)`B. `y=-asin(omegat+kx)`C. `y=asin(omegat-kx)`D. `y=-sin(omegat-kx)` |
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Answer» Correct Answer - B Here, stationary wave is formed, since node is formed at `x=0`. Therefore, the poing `x=0` is a fixed point `(i.e.,` it acts as a denser medium ) for a reflected wave. When reflectio of a wave takes place from denser medium, it is with a change in phase by `pi` rad. without any change in type of wave . The distance `x` of the particle will be taken as `-x` in the reflected wave with additional phase of `pi ` radian. Thus equation of reflected wave `y=asin (omegat=kx+pi)=-asin (omegat +kx)` |
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| 341. |
A stationary wave is described by the equation `y=10sin(x//3)cos(15pit),` where x is in cm and time t is in second. Calculate frequency and velocity of component waves. Also, find the amplitude of stationary waves at antinodes and find the position of first antinode. |
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Answer» The given eqn. is `y=10sin(x//3)cos15pit` Comparing with the standard form of eqn. `y=2sinkxcosomegat, ` we get `2r=10cm, k=(2ppi)/(lambda)=(1)/(3), omega=15pi` `lambda=6pi=6xx3.14=18.84cm` `omega =2piv=15pi,v=7.5Hz` `upsilon=vlambda=7.5xx18.84=141.3cms^(-1)` Amplitude of stationary wave `=10sin((x)/(3))` At antinode, amplitude is maximum, `sin((x)/(3))=max. :. (x)/(3)=pi//2 x=(3pi)/(2)=4.71cm` Amplitude of stationary wave at antinode `=10sin(x//3)=10xx1cm=0.1m` |
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| 342. |
The standing waves are set up by the superimposition of two waves: `y_(1)=0.05 sin (15pit-x)`and `y_(2)=0.05 sin (15pit+x)`, where x and y are in metre and t in seconds. Find the displacement of particle at `x=0.5m.` |
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Answer» According to superposition principle, resultant displacement of stationary wave is given by `y=y_(1)+y_(2)=0.05sin(15pit-x)` `+0.05sin(15pit+x)` `=0.05xx2sin((15pit-x+15pit+x)/(2))` `cos((15pit+x-15pit+x)/(2))=0.1cosxsin15pit` Amplitude of stationary wave, `2r=0.1cosx` At `x=0.5m,` `2r=0.1cos0.5=0.1cos((180^(@))/(pi))xx0.5` `=0.1 cos 28.65^(@)=0.1xx0.88=0.088m` |
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| 343. |
Statement-1`:` Two longitudinal waves given by equation `y_(1)(x,t)=2asin(omegat-kx)` and `y_(2)(x,t)=a sin (2 omegat-2kx)` will have equal intensity. Statement-2 `:` Intensity of waves of given frequency ini same medium is proportioal to square of amplitude only.A. Statement-1 is false, statement -2 is true.B. Statement-1 is true, statement-2 is false.C. Statement-1 is true, statement-2 is true, statement-2 is the correct explanation of statement-1.D. Statement-1 is true, statement-2 is true, statement-2 is not correct explanation of statement-1. |
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Answer» Correct Answer - b We know that intensity `(I)` of a wave. (i) is directly proportinoal to square of amplitude `(A)` (ii) is directlly proportional to square of angular frequency `(omega)` i.e., `Iprop A^(2)omega^(2)` For first wave, `I_(1)prop(2a)^(2)xxomega^(2)` or `I_(1)=k4a^(2)omega^(2)` For second wave, `I_(2) prop(a)^(2)xx(2omega)^(2)` or` I_(2)k4a^(2)omega^(2)` i.e., Intensity depends upon frequency also. Thus statement 1 is true but statment 2 is false. |
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| 344. |
Two standing bodies producing progressive waves are given by ` y_(1) = 4 sin 400 pi t and y_(2) = 3 sin 404 pi t` One of these bodies situated very near to the ears of a person who will hear : |
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Answer» The equation of progressive waves are: `y_(1)=4sin(400pit), and ` `y_(2)=3sin(404pit)` `:. a =4 and b=3` `omega_(1)=2pin_(1)=400pi,n_(1)=200Hz` `omega_(2)=2pin_(2)=404pi,n_(2)=202Hz` No. of beats `//` sec `=n_(2)-n_(1)=202-200=2` `(I_(max))/(I_(min))=((a+b)^(2))/((a-b)^(2))=((4+3)^(2))/((4-3)^(2))=(49)/(1)` |
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| 345. |
Which of the following conditions is not sufficient for SHM and why? (i) acceleration`prop` displacement, (ii) restoring force `prop` displacement. |
| Answer» Condition(i) is not sufficient because it gives no reference of the direction of acceleration, where as in SHM the acceleration is always in a direction opposite to that of the displacement . | |
| 346. |
A standing wave is represented by `y=2Asinkx cos omegat. ` If one of the component waves is `y_(1)=A sin (omega t-k x)`, what is the equation of the second component wave? |
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Answer» As, `2sinA cosB=sin(A+B)+sin(A-B)` `:. Y=2A sin k x cos omegat ` `=Asin (kx+omegat)+A sin(kx-omegat)` According to superposition principle, `y=y_(1)+y_(2),` and `y_(1)=A sin (omegat -kx)` `=-A sin(kx-omegat)` `:. Y_(2)=y-y_(1)` `=2Asin k x cos omega t +A sin (kx-omegat )` `=A sin (kx+omegat) +2A sin (kx-omegat)` `=A sin (kx+omegat)-2Asin(omegat-kx)` |
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| 347. |
A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force `F sin omegat `. If the amplitude of the particle maximum for `omega=omega_(2)` then `(omega_(0)` is the angular frequency of undampled oscillations ).A. `omega_(1)=omega_(0) ` and `omega_(2)!=omega_(0)`B. `omega_(1)=omega_(0) ` and `omega_(2)=omega_(0)`C. `omega_(1)!=omega_(0) ` and `omega_(2)=omega_(0)`D. `omega_(1)!=omega_(0) ` and `omega_(2)!=omega_(0)` |
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Answer» Correct Answer - C Amplitude resonance takes place at a frequency of external force which is less than the frequency of undampled vibrations. The velocity resosnace (max. energy) takes place when frequency of external periodic force is equal to the natural frequency of undamped vibrations. |
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| 348. |
Two sources of sound placed close to each other are wmitting progressive waves given by `y_1=4sin600pit` and `y_2=5sin608pit`. An observer located near these two sources of sound will hear:A. 4 beats per second with intensity ratio `25:16` between waxing and waningB. 8 beats per second with intensity ration `25:16` between waxing and waningC. 8 beats per second with intensity ration `81:1` between waxing and waningD. 4 beats per second with intensity ration` 81:1` between waxing and waning |
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Answer» Correct Answer - D Here, `y_(1)=4sin 600pit=a_(1)sin 2pin_(1)t` `:. a_(1)=4,n_(1)=(600pi)/(2pi)=300Hz` `y_(2)=5sin608pit=a_(2)sin 2pin_(2)t` `,. a_(2)=5,n_(2)=(608pi)/(2pi)=240Hz` Number of beats `//` sec `=n_(2)-n_(1)=304-300=4` `(I_(max))/(I_(min))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))=((4+5)^(2))/((4-5)^(2))=(81)/(1)` |
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| 349. |
Find the damping coefficient v of a sound wave if at distances `r_(1)=10m` and `r_(2)=20 m` from a point isotropic source of sound the sound wave intensity values differ by a factor `eta=4.5`. |
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Answer» Intensity of a spherical sound wave emitted from a point source in a homogeneous absorbing medium of wave damping coefficient `gamma` is given by `I=(1)/(2) rho a^(2) e^(-2gammar)omega^(2) v` So, Intensity of sound at a distance `r_(1)` from the source `(I_(1))/(r_(1)^(2))=(1//2rho a^(2) e^(-2gamma r_(1))omega^(2) v)/(r_(1)^(2))` and intensity of sound at a distance `r_(2)` from the source `=I_(2)//r_(2)^(2)=(1//2rhoa^(2)e^(-2gammar_(2))omega^(2)v)/(r_(2)^(2))` But according to the problem `(1)/(eta)(I_(1))/(r_(1)^(2))=(I_(2))/(r_(2)^(2))` So, `(etar_(1)^(2))/( r_(2)^(2))=e^(2 gamma(r_(2)-r_(1)))`or`1n(etar_(2)^(2))/(r_(1)^(2))=2 gamma(r_(2)-r_(1))` or,` gamma =(1n(etar_(2)^(2)//r_(1)^(2)))/(2(r_(2)-r_(1)))=6xx10^(-3)m^(-1)` |
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| 350. |
The displacement of a particle is represented by the equation `y=sin^(3)omegat.` The motion isA. non-periodicB. periodic but not simple harmonicC. simple harmonic with period `2pi//omega`D. simple harmonic with period `ppi//omega` |
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Answer» Correct Answer - B Given, `y=sin^(3)omegat=(1)/(4)[3sinomegat-sin3omegat]` As this motion is not represented by single harmonic function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion. |
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