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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.
| 101. |
Which of the following funchtions of time represent (a) periodic and (b) non-periodic motion? Give the period for each of period motion. `(omega` is any positive constant ) (i) `sinomegat+cosomegat` (ii) `sinomegat+cos2omegat+sin4omegat` (iii) `e^(-omegat)` (iv) `logomegat` |
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Answer» (i) `sinomegat+cosomegat` `=sqrt(2)[(1)/(sqrt(2))sinomegat+(1)/(sqrt(2))cosomegat]` `=sqrt(2)sin(omegat+pi//4)` As , `sqrt(2)sin(omegat+(pi)/(4))=sqrt(2)sin(omegat+(pi)/(4)+2pi)` so the given function is a periodic onoe and its period is `2pi//omega.` (ii)` sinomegat+cos2omegat+sin4omegat, ` it represents the periodic function with different angular frequency. Since, the period is the least time interval after which a function is repeated in its value. Here, `sinomegat,` has a period, `T=(2pi)/(omega),cos2omegat` has a period`(2pi)/(2omega)=(pi)/(omega)=(T)/(2)` and `sin4omegat` has a period ` (2pi)/(4omega)=(T)/(4)`. The last two terms repeat after any integral multiple of their period. Therefore, each term in the function repeats itself after time interval T. That is why, the given function is a periodic function iwth a period `T=2pi//omega` . (iii) The value of funtion `e^(-omegat)` , decreases with increasing time t and as `tprop oo`, it tends to zero. Therefore, the function is non-periodic. (iv) The value of function `logomegat` increases with time t. As `tprop oo`, `logomegat` approaches to `oo`. Therefore, the value of thhis function never repeates. Hence, it represents non-periodi function. |
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| 102. |
Solve the foregoing problem for the case of the pan having a mass `M`. Find the oscillation amplitude in this case. |
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Answer» Unlike the previous `(4.40)` problem the kinetic energy of body `m` decreases due to the perfelcty inelastic collision with the pan. Obviously the body `m` comes to strike the pan with velocity `v_(0)=sqrt(2gh)`. If `v` be the common velocity of the `'` body `m+` pena `"` system due to the collision thrn from the conservation of linear momentum `m v_(0)=(M+m)v` or `v=(m v_(0))/((M+m))=(msqrt(2gh))/((M+m)) ...(1)` At the moment the body `m` strikes the pan, the spring is compressed due to weight of the pan by the amount `Mg//k`. If `l` be the further compression fo the spring due to the velocity acquired by the `"` pan `-` body `m"` system, then from the conservatiion of mechanical energy of the said system in the field generated by the joint action of boty the gravity and spring forces `(1)/(2)(M+m)v^(2)+(M+m)gl=(1)/(2)k((Mg)/(k)+l)^(2)-(1)/(2)k((Mg)/(k))^(2)` or, `(1)/(2)(M+m)(m^(2)2gh)/((M+m))+(M+m)gl=(1)/(2)k((Mg)/(k))^(2)+(1)/(2)kl^(2)-Mgl-(1)/(2)k ((Mg)/(k))^(2)` (Using 1) or, `(1)/(2)kl^(2)-mgl-(m^(2)gh)/((m+M))=0` Thus` l=(mg+-sqrt(m^(2)g^(2)+(2kghm^(2))/(M+m)))/(k)` As minus sign is not acceptable `l=(mg)/(k)+(1)/(k)sqrt(m^(2)g^(2)+(2km^(2)gh)/((M+m)))` If the oscillating " pan `+` body `m"` system were at rest it correspong to their equilibrium position i.e. the spring were compressed by `((M+m)g)/(k)` therefore the amplitude of oscillation `a=l-(mg)/(k)=(mg)/(k)sqrt(1+(2hk)/(mg))` The mechanical energy of oscillation which is only conserved with the restoring forces becomes `E=U_("extreme")=(1)/(2)ka^(2)`( Because spring force is the only restoring force not the weight of the body ) Alternately `E=T_("mean")=(1)/(2)(M+m)a^(2)omega^(2)` thus `E=(1)/(2)(M+m)a^(2)((k)/(M+m))=(1)/(2)ka^(2)` |
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| 103. |
What type of mechanical waved do you expect to exist in (a) vacuum (b) air (c) inside the water (d) rock (e) on the surface of water ? |
| Answer» (a) No wave (b) longitudinal waves (c) longitudinal (d) transverse or longitudinal or both (separately ) (e) combined longitudinal and transverse (ripples ) | |
| 104. |
What provides the restoriing force for simple harmonic oscillations in the following cases? (i)simple pendulum (ii) spriing (iii) column of mercury in U tube. |
| Answer» The source of restoring force is as follows: (i) Simple pendulum: gravity (ii)sprinig: elasticity (iii) column mercury ini U tube: weight | |
| 105. |
Determine the period of oscillations of mercury of mass `m=200g` poured into a bent tube (figure) whose right arm forms an angle `theta=30^(@)` with vertical. The cross-sectional area of the tube is `S=0.50 cm^(2)`. The viscosity of mercury is to be neglected. |
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Answer» Let A be the uniform cross-sectional area of the tube. Let the level of liquid of density `rho` be depressed in the vertical arm of tube by distance y . The verticall rise of liquid in the slant portion of the tube `=ycostheta.` The vertical difference in liquid level in two arms of bent tuve `-y+ycostheta=(1+costheta)y` in two arms of bent tube will provide a restoring force to the liquid. So restoring force on the liquid is `F=-[` weight of the liquid column of height `(1+costheta(y]` `=-(1-costheta)yrhogA` `=-(1+costheta)Arhogy` ...(i) Since `Fpropy` and F is directed towards equilibrium position of liquid in two arms of bent tube, hence if the force is removed or liquid is left free, it will execute SHM in bent tube. Comparing (i), with , `F=-ky` We have, spring factore, `k=(1+costheta)Arhog.` Here, inertia factore `=` mass of mercury `=m=200g` Period of oscillation, `T=2pisqrt((i n ertia fact o r)/(spr i n g fact o r))` `=2xx(22)/70sqrt((200)/((1+cos30^(@))xx0.5xx13.6xx980))` `=0.8 s` |
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| 106. |
Can a motion be oscillatory but not simple harmonic? If your answer is yes give an example and if not explain why? |
| Answer» Yes, when a ball is dropped from a height on a perfectly elastic plane surface, the motino of ball is oscillatory but not simple harmonic as restoring force `F=mg=` constant and not `Fprop-y`. | |
| 107. |
What is the basic consition for th emotion of a particle to be SHM? |
| Answer» The necessary and sufficient condition for motino to be simple harmonic is that the restoring force must be linear, i.e., `F=-ky` or Torque,`tau=-ctheta` | |
| 108. |
Wat is a periodic wave function? |
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Answer» A wave fnction `y(x,t)` which satisfies that periodicity conditions of position and time is called a periodic wave function, i.e., (i) `y(x+mlambda,t)=y(x,t)` (ii) `y(x,t+nT)=y(x,t)` where `lambda=` wavelength of wave, `T=` period of the wave, n and m are integers. |
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| 109. |
What is a harmonic wave function ? |
| Answer» A harmonic wave function is a periodic function whose functional form is sine or cosine. | |
| 110. |
Two vibrating modes of an air column are shown in figure. What is the ratio of frequency of two notes ? |
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Answer» In mode (a), `l=lambda//2+lambda//4=3lambda//4, lambda=(4l)/(3)` `n_(1)=(upsilon)/(lambda)=(3upsilon)/(4l)` In mode (b), `l=lambda//2+lambda//2+(lambda)/(4)=(5lambda)/(4), lambda=(4l)/(5)` `n_(2)=(upsilon)/(lambda)=(5upsilon)/(4l) :. (n_(1))/(n_(2))=(3)/(5)` |
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| 111. |
An incident wave is represented by `y(x,t)=20sin(2x-4t)`. Write the expression for reflected wave (i) from a rigid boundary (ii) from an open boundary. |
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Answer» (i) The wave reflected from rigid boudnary is `y(x,t)=-20sin(-2x-4t)` `=20sin(2x+4t)` (ii) The wave reflected from open boundary is `y(x,t)=20sin(-2x-4t)` `=-20sin(2x+4t)`. |
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| 112. |
(i) For the wave on a string described in question, do all theh points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude ? Explain your answers. (ii) What is the amplitude of a point `0.375m` away from one end? |
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Answer» All the points on the string (i) have the same frequency except at the nodes (where frequency is zero) (ii) have the same phase everywhere in one loop except at the nodes. However, the amplitude of vibration at different points is different. From `y(x,t)=0.06sin((2pi)/(3))xcos(120pit),` the amplitude at `x(=0.375m) is =0.06 sin ((2pi)/(3))xx1=0.06 sin ((2pi)/(3))xx0.375=0.06sin((pi)/(4))=(0.06)/(sqrt(2))` `=0.042m` |
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| 113. |
A wave on a string is described by `y(x,t)=0.005sin(6.28x-314t)`, in which all quantities are in SI units. Calculate its amplitude and wavelength. |
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Answer» Here, `y(x,t)=0.005sin(6.28x-314t)` compare it with standart form `y(x,t)=rsin((2pi)/(lambda)x-(2pi)/(T)t)` `r=0.005m. ` This is the amplitude. `(2pi)/(lambda)=6.28, lambda=(2pi)/(6.28)=(2xx3.14)/(6.28)=1m` |
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| 114. |
Which of the following statements are true for the ocsillations of the mass suspended with a springA. time period varies directly as the square root of the suspended massB. The stiffer spring used show lesser time periodC. In the weightlessness state, the mass cannot execute oscillationsD. The system will have the same time period both on the moon and the earth. |
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Answer» Correct Answer - A::B::D Time period `T=2pisqrt((m)/(k))` ltbr. i.e., `T prop sqrt(m)` `Tprop (1)/(sqrt(k))` The time period is independent of `g`, hence `T` will be same on the moon and the earth. In the weightlessness state, the loaded spring one pulled and left will execute oscillations. |
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| 115. |
Determing whether or not th efollowing quantities can be in the same direction for a simple harmonic motion, (a) displacement and velocity (b) velocity and acceleration © displacement and acceleration. |
| Answer» (a)Yes, when the particle is moving from equilibrium position to extremen position (b) Yes, when the particle is moving from extreme position to mean position. (c) No, because in SHM the displacement is always opposite to acceleration. | |
| 116. |
Which of the following quantities are always zero in a S.H.M. ? Here, `vec(r),vec(upsilon)`and `vec(a)` represent amplitude, velocity and acceleration vectore.A. `vec(r)xxvec(upsilon)`B. `vec(r)xxvec(a)`C. `vec(a)xxvec(F)`D. `vec(r)xxvec(F)` |
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Answer» Correct Answer - A::B::C::D act along the same straight line either in the same direction of in opposite direction. So angle between each two vectors, `i.e., theta` is either zero of `180^(@0` and sin `180^(@0` both are zero. |
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| 117. |
Sound is produced at a time in two exactly identical strings, one of rubber and other of steel. In which string will the sound reach the other end earlier and why? |
| Answer» In steel string, sound will reach the other end earlier, because `(Y//rho)` is larger for steel. | |
| 118. |
What are harmonics? |
| Answer» Harmonics are the notes of frequencies which are equal to or integral multiple of the fundamental frequency. | |
| 119. |
What are fundamental note and overtones? |
| Answer» When a source is sounded, it generally vibrates in more than one mode and therefore, emits tones of different frequencies, the tone of lowest frequency is called the fundamental note and the tones of higher frequencies are called overtones. | |
| 120. |
Why sound can be hear more distinctly at a greater distance over water surface ? |
| Answer» This is because sound waves are almost totally reflected from the air water interface. The critical angle for air water interface is only `14^(@)`. | |
| 121. |
Air gets thinner as we go up in the atmosphere. Will the velocity of sound change? |
| Answer» As we move up, the pressure (P) of air and density of air `(rho)` , both decrease,. As `v=sqrt((gammaRT)/(M))` , therefore, velocity of sound will not change so long as temperture T of air remains constant. | |
| 122. |
Statement-1 `:` In under water communication, sound waves with low frequency are better than those with high frequency. Statement-2 `:` When sound pressure amplitude exceeds the atmospheric and hyrostatic pressure, it sets up eddy and turbulent motions in water.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 - c In under water communications, sound waves with higher frequency are preferred as they would carry more energy and can travel larger distances without dispersion. |
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| 123. |
The amplitude of a wave disturbance propagating in the positive x-direction is given by `y = (1)/((1 + x))^(2)` at time `t = 0` and by `y = (1)/([1+(x - 1)^(2)])` at `t = 2 seconds`, `x and y` are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is ............... m//s`.A. `0.2m//s`B. `0.5m//s`C. `0.3m//s`D. `0.4m//s` |
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Answer» Correct Answer - B Here, at `t=0, y=(1)/(1+x^(2))` or`1+x^(2)=(1)/(y)` `x^(2)=(1)/(y)-1=(1-y)/(y)` or `x=((1-y)/(y))^(1//2)` Also at `t=2s` `y=(1)/([1+(x-1)^(2)])` or `[1+(x-1)^(2)]=(1)/(y)` `(x-1)^(2)=(1)/(y)-1=(1-y)/(y)` or `(x-1)=((1-y)/(y))^(1//2)` `x=1+((1-y)/(y))^(1//2),`As `upsilon=(x_(2)-x_(1))/(t_(2)-t_(1))` Here, `x_(2)=[1+((1-y)/(y))^(1//2)]` and `x_(1)=((1-y)/(y))^(1//2)` `t^(2)-t^(1)=2-0=2s` `:. upsiklon=(1)/(2-0)=0.5ms^(-1)` |
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| 124. |
The waves produced by a motor boat sailing in water are :A. neither longitudinal nor transverseB. both, longitudinal and transverseC. only longitudinalD. only transverse |
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Answer» Correct Answer - B Water waves produced by a motor boat sailing in water are both, longitudinal and transverse. |
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| 125. |
Explain why waves on strings are always transverse? |
| Answer» A string is non stretchable i.e. compressions and rarefactions cannot be produced in strings. Therefore, longitudinal waves in strigs are not possible. Strings do have eleasticity of shape. Therefore, waves on strigns on strings are transverse. | |
| 126. |
Velocity of sound in air …………………… by ………………….. For every…………………… |
| Answer» increases, 0.61m`//`s, `1^(@)C` rise in temperature. | |
| 127. |
What is the effect of pressure on speed of sound in air? |
| Answer» Change is the effect of pressure on speed of sound in air, provided temperature remains constant. | |
| 128. |
Assertion`:` A vibrating tuning fork sounds louder when its stem is pressed against a desk top. Reason `:` When a wave reaches another denser medium, part of the wave is reflected.A. both, Assertion and Reason are true and Reason is the correct explanation of the Assertion.B. both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both,Assertion and Reason are false. |
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Answer» Correct Answer - B On pressing the stem agains a desk top, sound becomes louder, because area of vibration increases. The given reason though true is not a correct explanation of the assertion. |
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| 129. |
Assertion `:` The rotation of the earth about its axis is a S.H.M. Reason `:` The earth completes each roation about its axis in 24 hours while orbiting around earth.A. both, Assertion and Reason are true and Reason is the correct explanation of the Assertion.B. both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both,Assertion and Reason are false. |
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Answer» Correct Answer - d Here, both the assertion and reason are false. The rotation of earth about its axis is periodic but not to and fro about a fixed point, hence not a S.H.M. |
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| 130. |
Earthwuakes generate sound waves inside the earth. Unilke a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically, the speed of S wave is about 4.0 km `s^(-1)`, and that of P wave is 8.0 `kms^(-1)`. A seismograph records P and S waved from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, how far away does the earthquake occur? |
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Answer» Let `upsilon_(1), upsilon_(2)` be the velolcities of S waves P waves, and `t_(1), t_(2) ` be the time taken by these waves to travel to the position of seismograph. If l is the distance of occurrence of earthquake from the seismograph, then `l=upsilon_(1)t_(1)=upsilon_(2)t_(2)` ...(i) Now, `upsilon_(1)=4kms^(-1) and upsilon_(2)=8kms^(-1)` `:. 4t_(1)=8t_(2) or t_(1)=2t_(2)` ...(ii) Also, `t_(1)-t_(2)=4min=240s`. Usint (ii), `2t_(2)-t_(2)=240s, t_(2)=240s` `t_(1)=2t_(2)=2xx240=480s.` Now, from (i) `l=upsilon_(1)t_(1)=4xx480=1920km` Hence earthquake occurs 1920km away from the seismograph. |
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| 131. |
What determines the type of wave motion in a medium ? |
| Answer» Type of wave motion is determined by (i) nature of the medium (ii) mode of excitation of wave motion. | |
| 132. |
What are pressure waves? |
| Answer» Pressure waves are the longitudinal waves which involve changes in volume and pressure during their propagation through a medium. | |
| 133. |
Which property is common in all types of mechanical waves? |
| Answer» All mechanical waves travel in a medium, but the medium does not advance alongwith the wave. | |
| 134. |
What do mechanical waves transfer : energy, matter, both or neither ? |
| Answer» Mechanical waves transfer only energy from one point to another. | |
| 135. |
The propagation constant of a wave is also calle itsA. wave numberB. wavelengthC. frequencyD. angular wave number |
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Answer» Correct Answer - D angular wave number |
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| 136. |
What is a non dispersive medium? |
| Answer» A medium in which speed of a wave is independent of frequency of wave is called a non-dispersive medium. For example, air is a non-dispersive medium for sound waves. | |
| 137. |
Waves are generated on a water surface. Calculated the phase difference between two points A and B, when (i) A and B lie on the same wave front at a distance `=2lambda` between them. (ii) A and B lie on successive crests separated by 1 metre. (iii) A and B lie on successive troughs separated by 1.5m |
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Answer» (i) On the same wave front , phase diff. `=0`. Linear distance between the points does not matter. (ii) Between two successive crests, path diff. `=lambda` `:. `phase diff. `=2pi` radian. (iii) Again, path difference between two successive troughs `=lambda` `:. ` phase diff. between A and B `=2pi` radian |
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| 138. |
During earthquake, both longifudinal and transverse waves are produced having speeds `4.0km//s` and `8.0km//s` respectively. If the first transverse wave reahes the seismograph 8 miniutes after the arrival of first longitudinal wave, then find the distance of the position, where the earthquake occured. |
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Answer» Let d be the distance of the seismograph from the position of occurrence of earthquake. Tiem taken by longitudinal waves to travel this distance, `t_(1)=(d)/(upsilon_(1))=(d)/(4)s` Timem taken by transverse waves to travel lthis distance, `t_(2)=(d)/(upsilon_(2))=(d)/(8)s` As `t_(1)-t_(2)=8 m i n =8xx60s` `:. (d)/(4)-(d)/(8)=480 or (d)/(8)=480` `d=480xx8km=3840km` |
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| 139. |
A light pointer fixed to one prong of a tuning fork touches gnetly a smoked vertical plate. The fork is set vibrating and the plate is allowed to fall freely. 8 complete oscilllations are counted when the plate falls through 10cm.What is the frequency of the tuning fork? |
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Answer» Time taken by the plate to fall vertically through `h=10cm` is `t=sqrt((2h)/(g))=sqrt((2xx10)/(980))=(1)/(7)s` No. of vibrations recorded in `(1)/(7)s =8` `:.` No. of vibrations of fork in 1 s `=(8)/(1//7)` `=56Hz` |
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| 140. |
Assuming a particle to have the form of a ball and to absorb all incient light, find the radius of a particle for which its gravitational attraction to the sun is counterbalanced by the forces that light exerts on it. The power of light raiated by the sun equals `P = 4.10^(26)W`, and the density of the particle is `rho = 1.0 g//cm^(3)`. |
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Answer» Let `r =`radius of the ball `R =` distance between the ball & the sun `(rlt lt R)`. `M =` mass of the sun `gamma =` gravitational constant Then `(gammaM)/(R^(2)) (4pi)/(3)r^(3) rho = (P)/(4piR^(2)) pir^(2).(1)/(c )` (the factor `(1)/(c )` converts the enegry recived on the right into momentum recived. Then the rifht hand side is the momentum recived per unit time and must equal the negative of the impressed force for equilibrium). Thus `r = (3P)/(16pi gamma Mc rho) = 0.606 mu m`. |
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| 141. |
In linear SHM, a particle moves ……………………… near the mean position and ……………….. Near the extremen position. |
| Answer» Correct Answer - faster, slower | |
| 142. |
Which of the following relationship between the acceleration, a and the displacement x of a particle involve simple harmonic motion. (a) `a=7.0x` (b) `a=-200x^(2)` (c) `a=-10x` (d) `a=100x^(3)` |
| Answer» In S.H.M., acceleration a is related to displacement by the relation of the form `a=-kx`, which is for relation (c). | |
| 143. |
The frequency of total energy of a particle in SHM is ……………………. |
| Answer» Correct Answer - zero | |
| 144. |
The acceleration in SHM is always in ……………….. To that of displacement. |
| Answer» Correct Answer - opposite phase | |
| 145. |
A body weighing 20 g has a velcoity `8cms^(-1)` after one second of its starting from mean position. If the time period is 8 seconds, find the kinetic energy, potential energy and total energy. |
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Answer» Here, `m=20g, T=8s, omega=(2pi)/(T)=(2pi)/(8)=(pi)/(4)rad//s` When `t=1s, v=8cms^(-1)` Now, `v=Aomega cos omega t` `8=Axx(pi)/(4) cos ((pi)/(4)xx1)=A xx(pi)/(4)xx(1)/(sqrt(2))` or `A=(32sqrt(2))/(pi)` Total energy `=(1)/(2)momega^(2)A^(2)` `=(1)/(2)xx20xx((pi)/(4))^(2)xx((32sqrt(2))/(pi))^(2)=1280 ergs` Kinetic energy `=(1)/(2)mv^(2)=(1)/(2)xx20xx8^(2)=640erg` `PE= ` Total energy `-` Kinetic energy `=1280-640=640erg` |
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| 146. |
A block lying on a horizontal table executes SHM of period 1 second, horizontally. What is the maximum amplitude for which the block does not slide if coeff. Of friction between block and surface is `0.4, pi^(2)=10. ` |
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Answer» `A_(max)=omega^(2)a=mug` or `a=mug//omega^(2)=mug//(2pi//T)^(2)` |
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| 147. |
Maximum velocity of a particle in SHM is 10 cm`//`s . What is the average velocity during motion from one extreme position to other extreme position ? |
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Answer» Max. velocity, `v_(m)=aomega=a 2pi//T` …(i) Average velocity `=(t otal displacement)/(t otal time)` `:. v_(av)=(2a)/(T//2)=(4a)/(t)=4((v_(m))/(2pi))` [from (i)] `=(2v_(m))/(pi)=(2xx10)/(pi)=(20)/(pi)cm//s` |
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| 148. |
A body is vibrating with SHM of amplitude 15 cm and frequency 4 Hz. Compute the maximum values of the acceleration and velocity. |
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Answer» Max. value of acc. `=omega^(2)a=4pi^(2)v^(2)a` Max. value of velocity `=omega a =2piva` |
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| 149. |
When the displacement is one half of the amplitude, what fraction is potential in SHM ? At what displacement is the energy half kinetic and half potential? |
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Answer» Fraction of total energy which is kinetic `=(K.E.)/(T.E.)` `=((1)/(2)momega^(2)(a^(2)-y^(2)))/((1)/(2)m omega^(2)a^(2))=(a^(2)-y^(2))/(a^(2))=(a^(2)-(a^(2))/(4))/(a^(2))=(3)/(4)` Fraction of total energy which is potential `=(P.E.)/(T.E.)=((1)/(2)momega^(2)y^(2))/((1)/(2)momega^(2)a^(2))=(y^(2))/(a^(2))=(a^(2)//4)/(a^(2))=(1)/(4)` KE `=` PE `(1)/(2)momega^(2)(a^(2)-y^(2))=(1)/(2)momega^(2)y^(2)` On solving , `y=a//sqrt(2)` |
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| 150. |
A particle executes simple harmonic motion of amplitude A (i) At what distance from the extreme positino is its kinetic energy equal to half its potential energy ? (ii) At what position from the extreme position is its speed half the maximum speed. |
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Answer» If y is th edisplacement of particle from the mean position, then P.E. of particle , `U=(1)/(2)ky^(2)` and K.E. of particle, `K=(1)/(2)k (A^(2)-y^(2))` where A is the amplitude of oscillation and k is a force constant. As per question, `K=(U)/(2), so` `(1)/(2)k(A^(2)-y^(2))=(1)/(2)((1)/(2)ky^(2))` or `A^(2)-y^(2)=(y^(2))/(2) or 2A^(2)-2y^(2)=y^(2)` or `y=ssqrt((2)/(3))A` ` :.` Distnace of location from extreme position `=A-sqrt((2)/(3))A=A(sqrt(3)-sqrt(2))//sqrt(3)` `=A(1.732-1514)//1.732=0.83A` `=0.183` times the amplitude (ii) `v_(max)=Aomega` and speed at displacement from mean position is `v=omegasqrt(A^(2)-y^(2))` As per question, `v=v_(max)//2` As `omega sqrt(A^(2)-y^(2))=Aomega//2 or A^(2)-y^(2)=A^(2)//4` or `y^(2)=A^(2)-A^(2)=3A^(2)//4` or ` y =sqrt(3)A//2` `:.` Distance from the extreme position is `=A-y=A-(sqrt(3)A)/(2)=A[(2-sqrt(3))/(2)]` `=A[(2-1.732)/(2)]=0.134A` `=0.134` times the amplitude. |
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