

InterviewSolution
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.
451. |
If a simple pendulum is suspended in a lift and lift is moving downwards an acceleration, time period of simple pendulumA. increasesB. decreasesC. remains constantD. first increases and then decreases |
Answer» Correct Answer - A | |
452. |
Time period of simple pendulum suspended from a metallic wireA. increases with increases in temperature of the wireB. increases with decreasing temperature of the wireC. decreases with decreasing in temperatureD. can not be predicted |
Answer» Correct Answer - A | |
453. |
In `SHM` at the equilibrium position (i) displacement is minimum (ii) acceleration is zero (iii) velocity is maximum (iv) potential energy is maximumA. all are trueB. ii,iii,iv are trueC. ii,iii trueD. i,ii,iv true |
Answer» Correct Answer - C | |
454. |
Time period of simple pendulum of wire is independentA. mass of the bob and amplitude of oscillationsB. amplitude of oscillationC. temperature of the bobD. acceleration due to gravity |
Answer» Correct Answer - A | |
455. |
The amplitude of a simple harmonic oscillation is doubled. How does this affect (i) periodic time (ii) maximum velocity (iii) maximum acceleration and (iv) maximum energyA. ii & iiiB. i, ii & iiiC. i& iiiD. i, ii,iii & iv |
Answer» Correct Answer - A | |
456. |
The time period of a body executing SHM is 0.05 s. if the amplitude is 4 cm, then what will be the maximum velocity and maximum acceleration of the body? |
Answer» Correct Answer - `16pims^(-1),64pi^(2)ms^(-2)` |
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457. |
The maximum velocity of a particle executing SHM is `100cms^(-1)` and the maximum acceleration is `157cms^(-2)` determine the periodic time |
Answer» Correct Answer - 4 |
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458. |
The period of simple pendulumA. increases with decreases in lengthB. increases with increases in lengthC. increases with keeping length constantD. decrease with decrease in acceleration due to gravity |
Answer» Correct Answer - B | |
459. |
Time period of simple pendulum increase withA. increase in value of gB. increase in amplitudeC. decrease in value of gD. decrease in amplitude |
Answer» Correct Answer - C | |
460. |
The equation `T=2pisqrt(l//g)` is valid only whenA. the length of the pendulum is less than the radius the earthB. the length of the pendulum is greater than or equal to radius by the earthC. the length of pendulum is independent of the radius of the earthD. length of the pendulum is equal to radius and the earth. |
Answer» Correct Answer - A | |
461. |
Answer the following questions: (a) Time period of particle is S.H.M. depends on the force constant k and mass m of the particle `: T=2pisqrt(m//k).` A simple pendulum executes S.H.M. approximately. Why then is the time-period of a pendulum independent of the mass of the pendulum? (b) The motion of simple pendulum is approximately simple harmonic for small angles of oscillation. For large angle of oscillation, a more involved analysis (beyond the scope of this book) shows that T is greater that `2pisqrt(l//g)`. Think of a quantitative argument to appreciate this result. (c) A man with a wrist watch on his hand falls from the top of tower. Does the watch give correct time during the free fall? (d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity? |
Answer» (a) In case of a spring k does not depend upon m. However, in case of a simple pendulum, k is directly proportional to m and hence the ratio `(m)/(k)` is a constant quantity. (b). The restoring force for the bob of the pendulum is given by `F =- mg sin theta` If `theta` is small, then `sin theta = theta = (y)/(l) :. F =- (mg)/(l)y` i.e, the motion is simple harmonic and time period is `T = 2pi sqrt((l)/(g))`. Clearly, the above formula is obtained only if we apply the approximation `sin theta = theta`. For large angles, this approximation is not valid and T is greater than `2pi sqrt((l)/(g))`. (c) The wrist watch uses and electronic system or spring sysyem to give the time, which does not change with acceleration due to gravity. therefore, watch gives the correct time. (d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero i.e., the pendulum will not vibrate at all |
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462. |
The frequency of a vibrating body situated in air ……. (a) is the same as its natural frequency (b) is higher than its natural frequency(c) is lower than its natural frequency (d) can have any value |
Answer» (c) is lower than its natural frequency |
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463. |
A particle of mass `2g` is initially displaced through `2cm` and then released. The frictional force constant due to air on it is `12 xx 10^(-3)N//m`. The restoring force constant is `50 xx 10^(-3)N//m`. If it is in oscillatory motion, its time period isA. `pi sec`B. `pi//2 sec`C. `2pi sec`D. `4pi sec` |
Answer» Correct Answer - B `gamma = (b)/(2m), omega_(o) = sqrt((k)/(m))` and time period `T = (2pi)/(sqrt(omega_(o)^(2)-gamma^(2)))` |
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464. |
The velocities of a body executing S.H.M. are 3 cm/s and 4 cm/s when the displacements from the mean position are 4 cm and 3 cm, then the period of oscillation isA. `(1)/(2pi) s`B. `(x)/(2) s`C. `(2)/(x) s`D. `2pis` |
Answer» Correct Answer - D `T=2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))=2pisqrt((9-16)/(9-16))=2pi` |
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465. |
The ratio of instantaneous velocity and the average speed of the particle performing S.H.M. isA. `2pi`B. `(2)/(pi)`C. `(pi)/(2)`D. `(pi)/(2)sqrt(1-(x^(2))/(A^(2)))` |
Answer» Correct Answer - D `(v)/(v_(av))=(omegasqrt(A^(2)-x^(2)))/((2Aomega)/pi)` `=(pi)/(2)sqrt((A^(2)-x^(2))/(A^(2)))=(pi)/(2)sqrt(1-(x^(2))/(A^(2)))` |
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466. |
The ratio of instantaneous speed at centre to average speed of a particle performing S.H.M. isA. `2pi`B. `2//pi`C. `(pi)/(2)`D. `(pi)/(2)sqrt(1-(x^(2))/(A^(2)))` |
Answer» Correct Answer - C `(v_("max"))/(v_(av))=(Aomega)/((2Aomega)/(omega))=(pi)/(2)` |
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467. |
How will the period of a simple pendulum change when its length is doubled? |
Answer» Tnew = √2 Told. |
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468. |
Define non dissipative medium. |
Answer» A medium in which speed of wave motion is independent of frequency of wave is called non-dispersive medium. For sound, air is non dispersive medium. |
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469. |
What is Oscillation? |
Answer» To and for motion in the same path is called Oscillation. |
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470. |
The displacement x (in metre ) of a particle in, simple harmonic motion is related to time t ( in second ) as ` x= 0.01 cos (pi t + pi /4)` the frequency of the motion will beA. 0.5 HzB. 1 HzC. `pi//2 Hz`D. `pi Hz` |
Answer» Correct Answer - A `omega=pi` `therefore 2pin=pi therefore n=(1)/(2)=0.5` |
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471. |
What is the condition for motion of a particle to be SHM? |
Answer» Acceleration should be proportional to displacement and always directed towards mean position. |
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472. |
Which of the following condition is not sufficient for simple harmonic motion and why ? (i) acceleration and displacement (ii) restoring force and displacement |
Answer» Condition (i) is not sufficient, because direction of acceleration is not mentioned. In S.H.M, the acceleration is always in a direction opposite to that of the displacement. |
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473. |
The velocity of a particle performing S.H.M. are 0.13 m/s and 0.12 m/s. When it is at 0.12 m and 0.13 m from the mean position respectively. Then the amplitude isA. 0.117 mB. 0.177 mC. 11.7 mD. 10 m |
Answer» Correct Answer - B Tricks : `A=sqrt(v_(1)^(2)+v_(2)^(2))` |
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474. |
The equation of S.H.M. of a particle is `a+4pi^(2)x=0`, where a is instantaneous linear acceleration at displacement x. Then the frequency of motion isA. 1 HzB. `4pi Hz`C. `1//4 Hz`D. 4 Hz |
Answer» Correct Answer - A `a+4pi^(2)xx` comparing it with standard equation of S.H.M. `a+omega^(2)x` `omega^(2)=4pi` `omega=2pi` `2pin=2pitherefore n=1Hz` |
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475. |
What is the basic condition for the motion of a particle to have S.H.M.? |
Answer» The necessary and sufficient condition for motion to be S.H. is that the restoring force must be linear, i.e., F = -ky or torque, τ = -cθ. |
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476. |
How is acceleration related to the displacement in S.H. Motion? |
Answer» Acceleration ∝ (-displacement). |
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477. |
Define oscillatory motion. |
Answer» The motion which repeats itself over and over again about its mean position such that it remains confined within well defined limits (known as extreme positions) on either sides of the mean position, is called an oscillatory motion. |
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478. |
The bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminium? |
Answer» There will be no effect because the time period does not depend upon the nature of material of the bob. |
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479. |
How is S.H. oscillator related to circular motion? |
Answer» S.H.M. is the projection of uniform circular motion upon any of the diameters of the circle. |
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480. |
What is the displacement of a particle executing S.H.M.? |
Answer» It is the distance traveled by the particle from its mean position at ant instant. |
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481. |
What are harmonics? |
Answer» Harmonics are the notes of frequencies which are integral multiple of the fundamental frequency. |
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482. |
Define reverberation. |
Answer» Reverberation is the persistence of sound after the source has stopped to produce sound. |
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483. |
What do you mean by reverberation? What is reverberation time ? |
Answer» The phenomenon of persistence or prolongation of sound after the source has stopped emitting sound is called reverberation. The time for which the sound persists until it becomes inaudible is called the reverberation time. |
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484. |
Explain, why can we not hear an echo in a small room ? |
Answer» For an echo of a simple sound to be heard, the minimum distance between the speaker and the walls should be 17 m, so in any room having length less than 17 m, our ears can not distinguish between sound received directly and sound received after reflection. |
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485. |
Find an equation for the displacement velocity and acceleration of a particle executing S.H.M. |
Answer» Let a particle moves with constant angular speed ω along a circular path of radius 'a'. The foot of the perpendicular drawn from the position of the particle on the diameter of circular path executes S.H.M. When the particle moves from position P1 to P2 in time t, the displacement of Q (foot of the perpendicular drawn from the position of the particle on the diameter of the circle) executing SHM from ΔOP2 is given as or, OQ = OP2 sinθ y = a sinθ As θ = ωt The equation of the particle in SHM, y = a sin ωt and Velocity, v = dy/dt = d/dt (a sin ωt) or, v = aw cos ωt |
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486. |
Derive an expression for the total energy of a particle executing S.H.M. in terms of force constant (k). |
Answer» We know, potential energy of a particle executing S.H.M. (i.e., oscillator) at any instant is given by P.E. = 1/2 Mω2x2 ...(i) But ω = 2π/T = √{k/M} or, Mω2 = k ...(ii) Hence eqn. (i) becomes P.E. = 1/2 kx2 (iii) Also, kinetic energy of the particle executing S.H.m is given by K.E. = 1/2 Mω2(r2 - x2) where r = amplitude of the particle Using eqn. (ii), we get K.E. = 1/2 k(r2 - x2) ...(iv) Thus the total energy at any instant of the particle executing S.H.M. is given by E = K.E. + P.E. = 1/2 k(r2 - x2) + 1/2 kx2 or, E = 1/2 kr2 ....(v) |
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487. |
The time period of a simple pendulum of infinite length is (R=radius of earth).A. `T=2pisqrt((R)/(g))`B. `T=2pisqrt((R)/(2g))`C. `T=2pisqrt((2R)/(g))`D. T = infinite |
Answer» Correct Answer - A For infinitely long length of the pendulum, time period is, `T=2pisqrt((R)/(g))`. |
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488. |
Obtain an equation of a progressive wave. |
Answer» For a simple harmonic motion, displacement y = r sin(ωt + ϕ) Where r is amplitude, ω is angular velocity and ϕ is phase difference But ω = \(\frac{2\pi}{T}\) = 2πf And ϕ = \(\frac{2\pi}{T}x\) Where λ is wavelength and x is path difference. ∴ y = rsin 2π(\(\frac{t}{T}+\frac{x}{\lambda}\)) |
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489. |
If the length of a simple pendulum is equal to the radius of the earth, its time period will beA. `T= pi sqrt((R ) /(g))`B. `T= 2pi sqrt((2R)/(g))`C. `T= 2pi sqrt((R )/(g))`D. `T= 2pi sqrt((R )/(2g))` |
Answer» Correct Answer - 4 | |
490. |
The period of oscillation of a spring pendulum is T. If the spring is cut into four equal parts, then find the time period corresponding to each part.A. 1sB. 2sC. 3sD. 4s |
Answer» Correct Answer - 4 | |
491. |
Why the amplitude of the vibrating pendulum should be small? |
Answer» When amplitude of the vibrating pendulum is small then θ is small. Here the restoring force F = -mg sin θ = -mgθ = -mgx/l, where x is the displacement of bob and l is the length of pendulum. Hence F ∝ -x. Since F is directed towards mean position. Therefore, the motion of the bob of simple pendulum will be SHM, if θ is small. |
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492. |
When two waves of almost equal frequencies n1 and n2 reach at a point simultaneously, what is the time interval between successive maxima? |
Answer» Assume, n1 > n1 Frequency of Beat, vb = n1 - n2 ∴ Time period, \(T_b=\frac{1}{v_b}\) = \(\frac{1}{n_1-n_2}s\) |
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493. |
At what temperature (in ºC) will the speed of sound in air be 3 times its value at 0ºC? |
Answer» \(v\propto \sqrt T\) or \(\frac{v_r}{v_0}=\sqrt{\frac{T}{T_0}}\) From question, vT = 3v0 ∴ \(\frac{3v_0}{v_0}=\sqrt{{\frac{T}{273+0}}}\) or \(\sqrt T=3\sqrt{ 273}\) T = 9 × 273 = 2457 K T = 2457 – 273 T = 2184ºC |
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494. |
Two springs of force constants `1000N//m` and `2000N//m` are stretched by same force. The ratio of their respective potential enegries isA. `2:1`B. `1:2`C. `4:1`D. `1:4` |
Answer» Correct Answer - A `P.E. = (1)/(2)Kx^(2)` |
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495. |
Under what conditions does a sudden phase reversal of waves on reflection takes place ? |
Answer» On reflection from a denser medium, a wave suffers a sudden phase reversal. |
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496. |
Ratio of kinetic energy to potential energy of an oscillator when it is at distance `1//N` times of amplitude from mean positions inA. `n^(2)`B. `n^(2)+1`C. `1//n^(2)`D. `n^(2)-1` |
Answer» Correct Answer - D `(K.E.)/(P.E.)=((1)/(2)momega^(2)(A^(2)-x^(2)))/((1)/(2)momega^(2)x^(2))` `=(A^(2)-x^(2))/(x^(2))=(A^(2)-(A^(2))/(n^(2)))/((1)/(n^(2))A^(2))=n^(2)-1` |
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497. |
Ratio of kinetic energy to potential energy of an oscillator when it is at distance `1//N` times of amplitude from mean positions inA. `N^(2)+1`B. `N^(2)`C. `(1)/(N^(2))`D. `N^(2)-1` |
Answer» Correct Answer - D `(KE)/(PE) =(A^(2)-x^(2))/(x^(2))` |
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498. |
In an SHM, x is the displacement and a is the acceleration at time t. the plot of a against x for one complete oscillation will beA. a straight lineB. a circleC. an ellipseD. a sinusoidal curve |
Answer» Correct Answer - A In SHM, acceleration, `a=-omega^(2)x` `therefore`The graph between a and x is a straight line. |
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499. |
Sound waves from a point source are propagating in all directions. What will be the ratio of amplitudes at distances of x meter and y meter from the source ? |
Answer» Intensity = amplitude2 ∝ 1/(distance)2 . .. Required ratio = y/x |
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500. |
Two light spring of force constants `k_(1)` and `k_(2)` and a block of mass m are in one line `AB` on a smooth horizontal table such that one end of each spring is fixed on rigid supports and the other end is free as shown in the figure. The distance `CD` between the spring is `60cm`. If the block moves along `AB` with a velocity `120 cm//s` in between the springs, calculate the period of oscillation of the block. (take `k_(1) = 1.8 N//m`, `k_(2) = 3.2 N//m`, `m = 200 g`) A. `1.41 S`B. `2.81S`C. `5.64S`D. `1.92 S` |
Answer» Correct Answer - B Between `C` and `D` block will move with constant speed of `120cm//s`. Therefore, period of oscillation will be (starting from `C).T = t_(CD) +(T_(2))/(2) +t_(DC) +(T_(1))/(2)` Here, `T_(1) =2p sqrt((m)/(k_(1)))` and `T_(2) =2p sqrt(m)/(k_(2))`and `t_(CD) =t_(DC) = (60)/(120) =0.5s` `:. T = 0.5 +(2pi)/(2) sqrt((0.2)/(3.2)) +0.5 +(2pi)/(2) sqrt((0.2)/(1.8))` `(m = 200g =0.2 kg)T = 2.82 s` |
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