

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.
301. |
Is the motion of a simple pendulum strictly simple harmonic? |
Answer» It is not strictly simple harmonic because we make the assumption that Sinθ =θ, which is nearly valid only if θ is very small. |
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302. |
When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a isA. `a//4`B. `a//3`C. `a//2`D. `2a//3` |
Answer» Correct Answer - C Potential energy `= (1)/(2) m omega^(2)y^(2)=(1)/(4) xx (1)/(2) m omega^(2)a^(2)` `y^(2)=(a^(2))/(4)rArr y = +- (a)/(2)` |
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303. |
What happens to the period of a simple pendulum, if its length is doubled? What happens if the mass that is suspended is doubled? |
Answer» Time period of a simple pendulum is, T = 2π√{l/g} When length is doubled, the time period T' = √2 T Since, time period is independent of the suspended mass, hence with the increase in mass, the time period of the pendulum will remain the same. |
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304. |
Why all the stringed instruments are provided with hollow boxes? |
Answer» The stringed instrument are provided with hollow boxed in order to increase the surface area of vibration, which increases the loudness/ intensity (I) of the sound produces [As I ∝ a2 ]. Moreover, the air inside the hollow box is set into forces vibration which also increases the loudness of sound produced. |
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305. |
Can beats be observed in two light sources of nearly equal frequencies? |
Answer» No, the emission of light from atom is a random and rapid phenomena. The phase at a point due to two independent light sources will change rapidly and rapidly and randomly. Therefore, instead of beats, we shall get uniform intensity. However, if light sources are laser beams of nearly equal frequencies, we may be able to observe the phenomenon of beats in light. |
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306. |
The reverberation time of an empty hall is larger than that of a crowded hall. Why? |
Answer» The audience is also absorber of sound. The total absorption in case of crowded hall is more, therefore, as per Sabine formula, reverberation time of a crowded hall is smaller than that of an empty hall. |
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307. |
Define : (a) Time period(b) Frequency(c) Displacement |
Answer» (a) Time period : It is the least interval of time after which the periodic motion of a body repeats itself. S.l. units of time period is second. (b) Frequency : It is defined as the number of periodic motions executed by body per second. S.l unit of frequency is hertz (Hz). (c) Displacement: Its deviation from the mean position is called Displacement. |
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308. |
Explain the all types of Phases. |
Answer» It is a physical quantity, which completely express the position and direction of motion, of the particle at that instant with respect to its mean position. Y = a sin θ = a sin (ωt + φ0) here θ = ωt + φ0 = phase of vibrating particle. (i) Initial phase or epoch : It is the phase of a vibrating particle at t = 0. (ii) Same phase: Two vibrating particle are said to be in same phase, if the phase difference between them is an even multiple of n or path difference is an even multiple of (λ/2) or time interval is an even multiple of (T/2). (iii) Opposite phase : Opposite phase means the phase difference between the particle is an odd multiple of bar ω or the path difference is an odd multiple of λ or the time interval is an odd multiple of (T/2). (iv) Phase difference : If two particles performs S.H.M and their equation are y1 = a sin (ωt + φ1) and y2 = a sin (ωt + φ2) then phase difference ∆φ = (ωt + φ2) – (ωt + φ1) = φ2 – φ1 |
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309. |
If radius of a stretched wire is reduced to half, how is the wave speed affected? |
Answer» As \(v\propto\frac{1}{D}\) , therefore, wave speed becomes twice. |
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310. |
The velocity of sound air N.T.P. is 331 ms-1. Find its velocity when the temperature rises to 91ºC and its pressure is doubled. |
Answer» Here, v0 = 331 ms-1 Rise in temperature, t = 91ºC Since velocity of sound is not affected by the change in pressure, therefore we have to see the effect of temperature alone. As v1 = \(v_0\sqrt{\frac{273+t}{273}}= 331\sqrt{\frac{273+91}{273}}\) = 331\(\sqrt{1+\frac{91}{273}}\) = 331\(\sqrt{1+\frac{1}{3}}\) = 331 × \(\frac{2}{\sqrt 3}\) = 382.2 ms-1 |
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311. |
If tension of a wire is increased to four times, how is the wave speed changed? |
Answer» as \(v\propto\sqrt T\), therefore, wave speed becomes twice. |
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312. |
A mass of `0.98 kg` suspended using a spring of constant `K = 300 Nm^(-1)` is hit by a bullet of `20g` moving with a velocity `3m//s` vertically. The bullet gets embeded and oscillates with the mass. The amplitude of oscillation is (neglect gravity)A. `0.15 cm`B. `0.12 cm`C. `1.2 cm`D. `0.35 cm` |
Answer» Correct Answer - D Law conservation of linear momentum and `v = aomega` |
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313. |
A mass of `0.98 kg` suspended using a spring of constant `K = 300 Nm^(-1)` is hit by a bullet of `20g` moving with a velocity `3m//s` vertically. The bullet gets embeded and oscillates with the mass. The amplitude of oscillation is (neglect gravity)A. `0.15 cm`B. `0.35 cm`C. `1.2 cm`D. `12m` |
Answer» Correct Answer - B From `LCLM, (M+m)v = mu rarr (1)` From `LCE,(1)/(2)(M+m) v^(2) =(1)/(2)KA^(2)` |
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314. |
A particle executes S.H.M. of period 1.2 s and amplitude 8 cm. What is the time taken to travel 2.344 cm from the positive extremity?A. 0.17 sB. 1 sC. 0.15 sD. 0.7 s |
Answer» Correct Answer - C `A-2.344=Acosomegat` `cosomegat=(5.656)/(8)=0.707` `omegat=cos^(-1)(0.707)` `(2pi)/(T)t=(pi)/(4)` `t=0.15s` |
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315. |
The acceleration due to gravity on a planet is `3//2` times that on the earth. If length of a seconds pendulum on earth is `1m`, length of second pengulum on surface of planet isA. `0.7m`B. `1m`C. `1.7m`D. `1.5m` |
Answer» Correct Answer - D `sqrt((l)/(g) = constant` |
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316. |
In S.H.M., the velocity of the particle at the mean position is 1 m/s and acceleration at the extremity is `2 m//s^(2)`, the period of motion isA. 2 sB. 0.5 sC. 1 sD. 3.142 s |
Answer» Correct Answer - D `(a_(m))/(v_(m))=omega" "therefore omega=(2)/(1)=2` `T=(2pi)/(omega)=(2pi)/(2)=pim//s` `T=3.142s` |
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317. |
A spring when stretched upto 2 cm has potential energy u. If it is stretched upto 10 cm, potential energy would beA. u/25B. u/5C. 5 uD. 25 u |
Answer» Correct Answer - D `u_(1)=(1)/(2)kx_(1)^(2)andu_(2)=(1)/(2)kx_(2)^(2)` `(u_(2))/(u_(1))=(x_(2)^(2))/(x_(1)^(2))thereforeu_(2)(uxx100)/(4)=25u` |
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318. |
A spring has a natural length of 50 cm and a force constant of `2.0 xx 10^(3)` N`m^(-1)`. A body of mass 10 kg is suspended from it and the spring is stretched. If the body is pulled down to a length of 58 cm and released, it executes simple harmonic motion. The net force on the body when it is at its lowermost position of its oscillation is (10x) newton. Find value of x. (Take g=10m/`s^(2)`)A. 160 NB. 40 NC. 60 ND. 80 N |
Answer» Correct Answer - C `l=A+x+L` `58=A+50+x` `A+x=8` `F=Kx " "therefore x=(F)/(K)` `x=0.05m` `A=0.03m` |
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319. |
A particle in SHM has a period of 4s .It takes time `t_(1)` to start from mean position and reach half the amplitude. In another case it takes a time `t_(2)` to start from extreme position and reach half the amplitude. Find the ratio `t_(1)//t_(2)`A. `(t_(1))/(t_(2))`B. `(t_(1))/(t_(2))`C. `(t_(1))/(t_(2)) =2`D. `(t_(1))/(t_(2)) = 3` |
Answer» Correct Answer - B `(A)/(2) = A sin (omegat_(1)), (A)/(2) = A cos (omegat_(2))` |
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320. |
The length of a simple pendulum is increased by 1%. Its time period willA. increase by 0.5 %B. increase by 1 %C. increase by 2 %D. decrease by 0.5 % |
Answer» Correct Answer - A `(DeltaT)/(T)=(1)/(2)(Deltal)/(l)=(1)/(2)xx1%=0.5%` |
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321. |
A body has a time period `T_(1)` under the action of one force and `T_(2)` under the action of another force, the square of the time period when both the forces are acting in the same direction isA. `T_(1)^(2)T_(2)^(2)`B. `T_(1)^(2)//T_(2)^(2)`C. `T_(1)^(2) + T_(2)^(2)`D. `T_(1)^(2)T_(2)^(2)//(T_(1)^(2)+T_(2)^(2))` |
Answer» Correct Answer - D As `F_(1) = (m 4 pi^(2)a)/(pi^(2)) and F_(2) = (m 4 pi^(2)a)/(T_(2)^(2))` Net force, `F = F_(1) + F_(2) = 4 pi^(2) ma((1)/(T_(1)^(2))+(1)/(T_(2)^(2))) or (4 pi^(2)ma)/(T_(2))=4pi^(2)ma((1)/(T_(1)^(2))+(1)/(T_(2)^(2)))` or `(1)/(T^(2))=(1)/(T_(1)^(2))+(1)/(T_(2)^(2))" "rArr" "T^(2) = (T_(1)^(2)T_(2)^(2))/(T_(1)^(2)+T_(2)^(2))` |
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322. |
A body has a time period (4/5) s under the action of one force and (3/5) s under the action of another force. Then time period when both the forces are acting in the opposite direction simultaneously will beA. `4//5 s`B. `5//4 s`C. 1 sD. `12//5 sqrt(7) s` |
Answer» Correct Answer - D `F=F_(1)-F_(2)` `T^(2)prop(1)/(F) therefore Fprop(1)/(T^(2))` `T=(T_(1)T_(2))/(sqrt(T_(1)^(2)+T_(2)^(2)))=((4)/(5)xx(3)/(5))/(sqrt((16)/(25)-(9)/(25)))=((12)/(25))/((7)/(25))=(12)/(7)` |
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323. |
A body has a time period (4/5) s under the action of one force and (3/5) s under the action of another force. Then time period when both the forces are acting in the same direction simultaneously will beA. `4//5 s`B. `5//4 s`C. `1 s`D. `12//25 s` |
Answer» Correct Answer - D `F=F_(1)+F_(2)` `T^(2)prop(1)/(F) therefore Fprop(1)/(T^(2))` `T=(T_(1)T_(2))/(sqrt(T_(1)^(2)+T_(2)^(2)))=((4)/(5)xx(3)/(5))/(sqrt((16)/(25)+(9)/(25)))=(12//25)/(1)` `therefore T=0.48` |
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324. |
A body executing S.H.M. has a period of 3 s under one force and 4 s under another force. Then time period under the action of both the forces together in the same direction isA. 1.2 sB. 2.4 sC. 3.6 sD. 4.8 s |
Answer» Correct Answer - B `T=(T_(1)T_(2))/(sqrt(T_(1)^(2)+T_(2)^(2)))=(3xx4)/(sqrt(9xx16))=(3xx4)/(5)` `=(12)/(5)=2.4s` |
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325. |
Two simple harmonic motions are represented by `y_(1)=5 [sin 2 pi t + sqrt(3)cos 2 pi t] and y_(2) = 5 sin (2pit+(pi)/(4))` The ratio of their amplitudes isA. `1 : 1`B. `2 : 1`C. `1 : 3`D. `sqrt(3) : 1` |
Answer» Correct Answer - B `y_(1)=5[sin2pit+sqrt(3)cos 2pit]` `= 10((1)/(2)sin 2 pi + (sqrt(3))/(2)cos 2 pi t)` `=10("cos"(pi)/(3)sin 2 pi t + "sin"(pi)/(3)cos 2 pi t)` `=10[(sin 2 pi t + (pi)/(3))]rArr A_(1)=10` Similarly, `y_(2)=5sin(2pit+(pi)/(4))rArr A_(2)=5` Hence, `(A_(1))/(A_(2))=(10)/(5)=(2)/(1)` |
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326. |
A particle executes S.H.M., according to the displacement equation x = 6 sin `(3pit+pi//6)`m. Then the magnitude of its acceleration at t = 2 s isA. `3pi^(2)m//s^(2)`B. `9pi^(2)m//s^(2)`C. `18pi^(2)m//s^(2)`D. `27pi^(2)m//s^(2)` |
Answer» Correct Answer - D `x=6sin(3pit+(pi)/(6))` `x=6xxsin(3pixx2+(pi)/(6))` `=6sin((pi)/(6))=6xx(1)/(2)=3` `a=omega^(2)x` `=9pi^(2)xx3=27pi^(2)`. |
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327. |
If the displacement of a particle executing S.H.M. is given by x = 0.24 sin (400 t + 0.5)m, then the maximum velocity of the particle isA. `24 m//s`B. `48 m//s`C. `96 m//s`D. `72 m//s` |
Answer» Correct Answer - C `x=0.24sin(400t+0.5)m` `v_(m)=Aomega` `=0.24xx400=96m//s` |
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328. |
The displacement equation of a particle performing S.H.M. is x = 10 sin `(2pit+(pi)/(6))`m. Then the initial displacement of a particle isA. 5 mB. 2.5 mC. 0.5 mD. 0.25 m |
Answer» Correct Answer - A `x=10sin(2pit+(pi)/(6))` `=10sin(2pixx0+(pi)/(6))` `=10sin.(pi)/(6)=10xx(1)/(2)=5m` |
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329. |
A simple harmonic oscillator has its displacement varying as x = 15 sin `(2pit+pi//4)` metre. Then its initial displacement isA. 15 mB. `15//sqrt(2) m`C. `15 sqrt(2) m`D. `7.5 m` |
Answer» Correct Answer - B `x=15sin(2pit+(pi)/(4))` `=15sin(2pixx0+(pi)/(4))` `=15sin.(pi)/(4)` `=15xx(1)/(sqrt(2))=(15)/(sqrt(2))m` |
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330. |
A particle executing S.H.M. is given by `x=10sin(8t+pi//3)m`. Its velocity when it is at a distance 6 m from the mean position is,A. 36 m/sB. 10 m/sC. 80 m/sD. 64 m/s |
Answer» Correct Answer - D `x=10sin(8t+(pi)/(3))` `v=omegasqrt(A^(2)-x^(2))` `=8sqrt(100-36)=8sqrt(04)=8xx8=64m//s` |
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331. |
Which of following is charecteristic of `SHM`?A. Projection of unifrom circular motion on any diameterB. Periodic in natureC. Displacement time graph is sine curve.D. Acceleration is zero at the mean position |
Answer» Correct Answer - B | |
332. |
In `SHM` there is always a constant ratio between displacement of the body and itsA. velocityB. accelerationC. mass of the particleD. all of the above |
Answer» Correct Answer - B | |
333. |
A particle move along y-axis according to equation `y +3+4 cos omega t`. The motion of the particle isA. circularB. periodic but not `SHM`C. `SHM`D. oscillatory but not `SHM` |
Answer» Correct Answer - C | |
334. |
A particle moves on the X-axis according to the equation `x=x_0 sin^2omegat`. The motion simple harmonicA. with amplitude `x_(0)//2`B. with amplitude `2x_(0)`C. with time period `(pi)/(omega)`D. with time period `(pi)/(omega)` |
Answer» Correct Answer - C | |
335. |
Simple harmonic oscillations areA. one dimensionalB. three dimensionalC. two dimensionalD. four dimensional |
Answer» Correct Answer - A | |
336. |
A ball of mass 5 kg hanging from a spring oscillates with a time period of `2 pi s`. At any instant the ball is at equilibrium position, now the ball is removed, then spring shortens byA. `2 pi m`B. `9//2 m`C. `9.8 m`D. 2 m |
Answer» Correct Answer - C `T=2pisqrt((x)/(g))` |
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337. |
A block on a horizontal slab is moving horizontally with a simple harmonic motion of frequency two oscillations per second. If the coefficient of static friction between block and slab is 0.5, then the amplitude of the oscillation will be (if the block does not slip along the slab)A. 3.3 cmB. 3.5 cmC. 3.15 cmD. 7 cm |
Answer» Correct Answer - C `KA=mumg` `momega^(2)A=mumg` `A=(0.5xx10)/(4pi^(2)4)=3.15cm` |
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338. |
The velocity at mean position and acceleration at the extreme position have the phase difference ofA. `pi//2` radB. `pi` radC. `pi//4` radD. `3pi//4` rad |
Answer» Correct Answer - B The velocity and acceleration have a phase of `(pi)/(2)`. The mean and extreme position have a phase of `(pi)/(2)`, therefore a phase of `pi`. |
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339. |
What is the relation between uniform circular motion and S.H.N? |
Answer» Uniform form circular motion can be thought of as two simple harmonic motion operating at right angle to each other. |
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340. |
In S.H.M.,A. epoch and phase continuously changes with timeB. epoch and phase remains constant at all timesC. epoch remains constant while phase changes continuously with timeD. phase remains constant while epoch change continuously with time |
Answer» Correct Answer - C | |
341. |
The phase of a particle performing S.H.M. when the particle is at a distance of amplitude from mean position isA. `pi//2`B. `pi`C. `3pi//2`D. odd multiple of `pi//2` |
Answer» Correct Answer - D | |
342. |
The displacement of a particle performing S.H.M is `x=A" "sin(omegat+prop)`. The quantity `prop` is calledA. phase constantB. epochC. initial phaseD. all of the above |
Answer» Correct Answer - D | |
343. |
Which of the following is not simple harmonic function ?A. `y =a sin 2 omegat +b cos omega t`B. `y = a sin 2 omegat +b cos omegat`C. `y = (a^(2) +b^(2))^(½) sin omega t cos omega t`D. `y = 1-2 sin^(2) omegat` |
Answer» Correct Answer - B | |
344. |
A particle is moving in a circle with uniform speed its motion isA. periodic and simple harmonic.B. periodic but not simple harmonicC. non periodicD. None of the above |
Answer» Correct Answer - B | |
345. |
The function sin^(2) (omegat) represents.A. periodic but not simple harmonic motionB. simple harmonic motion with a period `(pi)/(omega)`C. simple harmonic motion with a period `(2pi)/(omega)`D. simple harmonic motion with a period `(4pi)/(omega)` |
Answer» Correct Answer - A | |
346. |
In given statements correct alternative isA. `SHM` is necessarily periodicB. `SHM` is necessarily oscillatoryC. oscillatory motion may be periodic.D. all of the above |
Answer» Correct Answer - D | |
347. |
A uniform plank of mass `m = 1kg`, free to move in the horizontal direction only, is placed at the top of a solid cylinder of mass `2m` and radius `R`. The plank is attched to a fixed wall by means of light spring of spring constant `k = 7N//m^(2)`. There is no slipping between the cylinder and the plank, and between the cylinder and the ground. the angular frequency of small oscillations of the system is |
Answer» Correct Answer - 2 Suppose that the plank is displaced from its equilibrium position by `x` at time `t`, the centre of the cylinder is, therefore, displaced by `x//2`. `:.` the mechanical energy of the system is given by, `K = K.E. ("Plank") +P.E. ("spring") +K.E. ("cylinder")` `E = (1)/(2)m((dx)/(dt))^(2) +(1)/(2)kx^(2) +(1)/(2)2m {(d)/(dt)((x)/(2))}^(2)` `+(1)/(2) ((1)/(2)2m.R^(2)) {(1)/(R)(d)/(dt)((x)/(2))}^(2) = (1)/(2) ((7)/(4)m)((dx)/(dt))^(2)+(1)/(2)kx^(2)` After differentiating the total energy and equating it to zero, one finds acceleration `=- omega^(2)x` The angular frequency, `omega = sqrt((4k)/(7m)) = 2 rad//sec` |
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348. |
Define Beats. |
Answer» When two sound waves of slightly different frequencies, travelling in a medium along the same direction, superimpose on each other, the intensity of the resultant sound at a particular position rises and falls regularly with time. This phenomenon is called beats. |
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349. |
Define Beat period and Beat frequency. |
Answer» Beat period : The time interval between two successive beats (i.e. two successive maxima of sound) is called beat period. Beat frequency : The number of beats produced per second is called beat frequency. |
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350. |
How will the time period of a simple pendulum change when its length is doubled ? |
Answer» Time period √2 times, as T α √l |
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