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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.
| 901. |
The relation between acceleration and displacement of four partical are given below Which, one of the particle is exempting simple harmonic motion ?A. `a_(x)=+2_(x)`B. `a_(x)=+2_(x)^(2)`C. `a_(x)=-2_(x)^(2)`D. `a_(x)=-2_(x)` |
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Answer» For motion to be be SHM acceleration of the particale must be proportional to negative of displacement. i.e., `aprop-(y" or "x)` We should be clear that y has to be linear. |
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| 902. |
The relation between acceleration and displacement of four particles are given below:(a) ax = + 2x.(b) ax = + 2x2.(c) ax = – 2x2.(d) ax = – 2x.Which one of the particles is executing simple harmonic motion? |
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Answer» (d) ax = – 2x. particles is executing simple harmonic motion |
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| 903. |
The displacement of a particle is represented by the equation y = sin3ωt . The motion is(a) non-periodic.(b) periodic but not simple harmonic.(c) simple harmonic with period 2π/ω.(d) simple harmonic with period π/ω. |
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Answer» (b) periodic but not simple harmonic |
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| 904. |
The displacement of a particle is represented by the equation y = 3 cos ((π/4)-2ωt). The motion of thw particle is(a) simple harmonic with period 2p/w.(b) simple harmonic with period π/ω.(c) periodic but not simple harmonic.(d) non periodic. |
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Answer» (b) Simple harmonic with period π/ω. |
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| 905. |
Two linear SHM of equal amplitudes `A` and frequencies `omega` and `2omega` are impressed on a particle along `x` and `y - axes` respectively. If the initial phase difference between them is `pi//2`. Find the resultant path followed by the particle.A. a circleB. a straight lineC. an ellipseD. a parabola |
| Answer» Correct Answer - A | |
| 906. |
If the displacement of a particle executing SHM is given by `y=0.30 sin (220t +0.64)` in metre , then the frequency and maximum velocity of the particle isA. 35 Hz, 66 m/sB. 45 Hz, 66 m/sC. 58 Hz, 113 m/sD. 35 Hz, 132 m/s |
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Answer» Correct Answer - A `x=0.30sin(220t+0.64)m` `omega=220` `2pin=220` `n=(220)/(2xx(22)/(7))=(10xx7)/(2)=35Hz` `v_(m)=Aomega` `=0.30xx220=66m//s` |
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| 907. |
The displacement of a particle executing SHM is given by `y=0.2sin50pi(t+0.01)`metre Calculate the amplitude,t he period, maximum velocity and the displacement at the start of the motion. |
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Answer» Correct Answer - 0.2m, 0.04s, `31.4ms^(-1),02m` |
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| 908. |
The motion of a 100 g mass tied to a spring is described by the equation `x=25cos(3t+(pi)/(4))cm`. Find (i) the angular velocity `omega`(ii) frquency v (iii) the time period T (iv) the force constant k (v) the amplitude A and (v) the phase angle |
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Answer» Correct Answer - (i). `omega=3rads^(-1),(ii)v=0.477Hz,(iii)T=2.06,(iv)k=0.9Nm^(-1),(v)A=0.25m, (vi) phi_(0)=pi//4rad` |
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| 909. |
Define :(a) Free oscillations(b) Damped oscillations(c) Forced oscillations |
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Answer» (a) Free oscillations: If a body, capable of oscillation, is slightly displaced from its position of equilibrium and then released, its starts oscillating with a frequency of its own. Such oscillations are called free oscillations. The frequency with which a body oscillates is called natural frequency. (b) Damped oscillations: The oscillations in which amplitude decreases gradually with the passage of time are called damped oscillations. (c) Forced oscillations: When a body oscillates under the influence of an external periodic force, not with its own natural frequency but the frequency of the external periodic force, its oscillations are said to be forced oscillations. |
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| 910. |
In a sound wave, a displacement node is a pressure antinode and vice- versa. Explain, why ? |
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Answer» At the point, where a compression and a rarefaction meet, the displacement is minimum and it is called displacement node. At this point, pressure difference is maximum i.e. at the same point it is a pressure antinode. On the other hand, at the mid point of compression or a rarefaction, the displacement variation is maximum i.e. such a point is pressure node, as pressure variation is minimum at such point. |
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| 911. |
What is the distance between a compression and its nearest rarefaction in a longitudinal wave? |
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Answer» Distance between a compression and adjoining rarefaction is γ/2. |
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| 912. |
A bob of simple pendulum of mass 1g is oscillating with a frequency 5 vibrations per second and its amplitude is 3cm. Find the kinetic energy and kinetic energy at that instant. |
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Answer» Correct Answer - 4441.5 erg |
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| 913. |
A pendulum 1 m long makes 20 vibrations in 40s. Find the time taken to make 30 vibrations, if its length is increased to 4 m. |
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Answer» Correct Answer - 120s |
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| 914. |
Ocean waves hitting a beach are always found to be nearly normal to the shore. Why? |
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Answer» Ocean waves are transverse waves travelling in concentric circles of ever increasing radii. When they hit the shore, their radius of curvature is so large that they can be treated as plane waves. Hence thy hit the shore nearly normal to the shore. |
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| 915. |
Define wave number and angular wave number and give their S.I. units. |
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Answer» Wave number is the number of waves present in a unit distance of medium. (v = 1/ λ) S.I. unit of k is rad m–1. Angular wave number or propagation constant is 2π/λ. It represents phase change per unit path difference and denoted by k = 2π/λ. S.I. unit of k is rad m–1. |
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| 916. |
The length of a simple pendulum is increased by 44%. The percentage increase in its time period will beA. 0.44B. `sqrt(44) %`C. 0.2D. 0.1 |
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Answer» Correct Answer - C `(T_(2))/(T_(1))=sqrt(1.44)=(T_(2))/(T_(1))=1.20` `(T_(2)-T_(1))/(T_(1))=0.20 therefore (T_(2)-T_(1))/(T_(1))%=20%` |
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| 917. |
The length of simple pendulum executing S.H.M is increased by 21%. By what % time period of pendulum increase ? |
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Answer» When a number of waves travel through the same region at the same time, each wave travels independently as if all other waves were absent. This characteristic of wave is known as independent behaviour of waves. For example we can distinguish different sounds in a full orchestra. |
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| 918. |
What are isochronous vibrations? |
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Answer» When the time period is independent of amplitude such an oscillation is called isochronous. |
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| 919. |
If the length of a simple pendulum is increased by 45% what is the percentage increase in its time period? |
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Answer» Correct Answer - 0.225 |
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| 920. |
A simple pendulum Is inside a spacecraft. What should be its time period vibration? |
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Answer» Pendulum does not oscillate. |
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| 921. |
The time taken by a simple pendulum to perform 100 vibration is 8 minutes 9 sec in bombay and 8 minutes 20 sec. in pune. Calcualte the ratio of acceleration due to gravity in bombay and pune. |
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Answer» Correct Answer - 1.0455 |
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| 922. |
A spring of force constant k is cut into two pieces whose lengths are in the ratio 1:2. The force constant of the longer piece?A. `k//2`B. `3k//2`C. `2k`D. `3k` |
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Answer» Correct Answer - B If K is the spring constant of the spring then `K_(1)` spring constant of longer parts of the spring is `K_(1)=(K)/(n)(n+1)=(3K)/(2)` |
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| 923. |
A light vertical spring is stretched by `0.2 cm` when a weight of `10g` is attached to its free end. The weight is further pulled down by `1cm` and released. Compute the frequency and maximum velocity of load. |
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Answer» (i) Force constant of the spring `K = ("Restoring Force")/("Increase in length") = (mg)/("Increase in length")` `= (10^(-2)x9.8)/(2x10^(-3)) = 49 N m^(-1)` Frequency `f = (1)/(2pi) sqrt((K)/(m)) = (1)/(2pi) sqrt((49)/(10^(-2))) = (35)/(pi) Hz` (ii) amplitude of motion `(A) =` distance through which the weight is further pulled down `= 1cm` `V_(max) = A omega = 10^(-2) m x 70 rads^(-1) = 0.7 ms^(-1)` |
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| 924. |
The periodic time of a mass suspended by a spring (force constant k) is T. if the spring is cut in three equal pieces, what will be the force constant of each part? If the same mass be suspended from one piece, what will be the periodic time? |
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Answer» Correct Answer - 3k, `T//sqrt(3)` |
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| 925. |
Two springs are jonied and connected to a mass m as shown in figure. If the force constants of the two springs are `k_(1) and k_(2)`, shown that frequency of oscillation of mass m is |
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Answer» Correct Answer - `v=(1)/(2pi)sqrt((k_(1)k_(2))/((k_(1)+k_(2))m))` |
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| 926. |
If the two particles performs S.H.M. of same initial phase angle but different amplitudes of individuals, then the resultant motion initial phase angle depends onA. initial phase angle onlyB. initial phase angle and amplitude of individualC. amplitude of individual onlyD. neither amplitude nor initial phase angle |
| Answer» Correct Answer - A | |
| 927. |
The force constant of a weightless spring is `16 N m^(-1)`. A body of mass `1.0 kg` suspended from it is pulled down through `5 cm` and then released. The maximum energy of the system (spring + body) will be |
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Answer» Correct Answer - (i). 1.57s (ii) `2xx10^(-2)J` |
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| 928. |
Spring is pulled down by 2 cm. What is amplitude of its motion?A. 0 cmB. 6 cmC. 2 cmD. 4 cm |
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Answer» Correct Answer - C `A=2cm` . |
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| 929. |
The time period of oscillation of a particle that executes `SHM` is `1.2s`. The time starting from mean position at which its velocity will be half of its velocity at mean position isA. `0.1s`B. `0.2s`C. `0.4s`D. `0.6s` |
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Answer» Correct Answer - A `V = (V_(max))/(2) rArr x =+-(sqrt(3)A)/(2), x = A cos ((2pi)/(T)t)` |
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| 930. |
Kinetic energy of the particle performing S.H.M. isA. harmonic motion and oscillatoryB. periodic motion but not oscillatoryC. oscillatory motion but not periodicD. periodic and oscillatory motion |
| Answer» Correct Answer - B | |
| 931. |
A body is performing S.H.M. Then its(a) average total energy per cycle is equal to its maximum kinetic energy.(b) average kinetic energy per cycle is equal to half of its maximum kinetic energy.(c) mean velocity over a complete cycle is equal to 2/π times of its maximum velocity.(d) root mean square velocity is 1/√2 times of its maximum velocity. |
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Answer» (a) average total energy per cycle is equal to its maximum kinetic energy. (b) average kinetic energy per cycle is equal to half of its maximum kinetic energy. (d) root mean square velocity is 1/√2 times of its maximum velocity. |
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| 932. |
The amplitude of a damped oscillator becomes `(1)/(27)^(th)` of its initial value after `6` minutes. Its amplitude after `2` minutes isA. `(A_(0))/(3)`B. `(A_(0))/(9)`C. `(A_(0))/(54)`D. `(A_(0))/(81)` |
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Answer» Correct Answer - A For damped oscillator `A = A_(0)^(-bt)` Here `(A_(0))/(27) = A_(0)e^(-6b), A_(2) = A_(0)e^(-2b) = A_(0) [e^(-6b)]^((1)/(3))` |
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| 933. |
The amplitude of damped oscillator becomes `1/3` in `2s`. Its amplitude after `6s` is `1//n` times the original. The value of `n` isA. `2^(3)`B. `3^(2)`C. `3^(1//3)`D. `3^(3)` |
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Answer» Correct Answer - D `A = A_(0)e^(-gammat)` |
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| 934. |
The total energy of a particle vibrating in SHM is proportional to the square of its ……..(a) velocity (b) acceleration (c) amplitude (d) none of these |
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Answer» (a) veloctiy |
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| 935. |
The amplitude of a vibrating body situated in a resisting medium ………(a) decreases linearly with time(b) decrease exponentially with time (c) decreases with time in some other manner (d) remains constant with time |
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Answer» (b) decreases exponentially with time |
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| 936. |
A mass `m = 8 kg` is attached to a spring passing over a pulled whose other end is fixed to ground and held in position so that the spring remains unstretched. The spring constant is `200 N//m`. The mass `m` is then released and begins to undergo small oscillations. Find maximum velocity of mass |
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Answer» Mean position will be at `Kx = mg` or `x = (mg)/(K) = (8 xx 10)/(200) =(2)/(5) = 0.4 m` This is also the amplitude of oscillation `A = 0.4m` `V_(max) = A• = A sqrt((K)/(m)) = (0.4) sqrt((200)/(8)) = 2m//s` |
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| 937. |
Frequency of a particle executing `SHM` is `10Hz`. The particle is suspended form a vertical spring. At the highest point of its oscillation the spring is unstretched. Find the maximum speed of the particle: `(g = 10 m//s^(2))` |
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Answer» Mean position of the particles is `(mg)/(K)` distance below unstretched position of spring. Therefore, amplitude of oscillation is `A = (mg)/(K) • = sqrt((K)/(m)) = 2•f = 20•` `:. (m)/(K) = (1)/(•^(2)), A = (g)/(•^(2))` Therefore, the maximum speed of particle will be `V_(max) = A• = (g)/(•^(2)) xx • =(g)/(•) = (1)/(2•)m//s` |
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| 938. |
What will be the speed of sound in perfectly rigid rod? |
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Answer» The speed of sound in a perfectly rigid rod will be infinite. |
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| 939. |
The time period of oscillation of a `SHO` is `(pi)/(2)s`. Its acceleration at a phase angle `(pi)/(3) rad` from exterme position is `2ms^(-1)`. What is its velocity at a displacement equal to half of its amplitude form mean position? (in `ms^(-1)`A. `0.707`B. `0.866`C. `sqrt(2)`D. `sqrt(3)` |
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Answer» Correct Answer - C `V = omega sqrt(A^(2) - x^(2)), a = omega^(2)x` |
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| 940. |
If the maximum speed of a SHO is `pi ms^(-1)`. Its average speed during one oscillations isA. `(pi)/(2)ms^(-1)`B. `(pi)/(4)ms^(-1)`C. `pims^(-1)`D. `2ms^(-1)` |
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Answer» Correct Answer - D `V_(m)=Aomega=pi,pi=AomegaimpliesA=(pi)/(omega)` Average speed `=(4A)/(T)=(4)/(T)xx(pi)/(omega)=(4omega)/(2pi)xx(pi)/(omega)` `=2ms^(-1)` |
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| 941. |
A `SHO` has amplitude `A` and time period `T`. The maximum velocity will beA. `4AT`B. `(2A)/(T)`C. `2pisqrt(A//T)`D. `2pi A//T` |
| Answer» Correct Answer - D | |
| 942. |
A particle performs SHM with a period `T` and amplitude a. The mean velocity of particle over the time interval during which it travels `a//2` from the extreme position isA. `(A)/(2T)`B. `(A)/(T)`C. `(2A)/(T)`D. `(3A)/(T)` |
| Answer» Correct Answer - D | |
| 943. |
The graph between instantaneous velocity and acceleration of a particle performing S.H.M. isA. parabolaB. straight lineC. ellipseD. circle |
| Answer» Correct Answer - C | |
| 944. |
The displacement of a particle of mass 1 kg in S.H.M. is x = 2 sin `(pit+phi)`m. Then variation of its PE in joule isA. `U=4pi^(2)sin^(2)(pit+phi)`B. `U=2pi^(2)sin^(2)(pit+phi)`C. `U=2pi^(2)cos^(2)(pit+phi)`D. `U=4pi^(2)cos^(2)(pit+phi)` |
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Answer» Correct Answer - B `P.E.=(1)/(2)kx^(2)=(1)/(2)momega^(2)4sin^(2)(pit+phi)` `=pi^(2)2sin^(2)(pit+phi)=2pi^(2)sin^(2)(pit+phi)` |
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| 945. |
The displacement of `SHO` is, `y = 6 sin(pit +pi//3)` find (1) Instants at which `PE` is minimum (or) `KE` is maximum. (2) Instants at which `PE` is maximum (or) `KE` is minimum. |
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Answer» `PE` is minimum (or) `KE` is maximum when `SHO` is at mean position i.e., `y =0` `y = 0 = 6 sin (pit +pi//3)` `rArr pi(t+(1)/(3)) =npi` here `n =1,2,3….` `t = n-(1)/(3)` `PE` is maximum (or) `KE` is minimum at `y = +-A` When `y = +- 6` i.e `sin (pit + pi//3) = +1` `(pit+(pi)/(3)) =(2n+1) (pi)/(2)` here `n = 0,1,2,3...` `t = (1)/(3) =(2n+1)/(2)(s), t = n+(1)/(2)-(1)/(3) (s)` |
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| 946. |
The graph between instantaneous velocity and acceleration of a particle performing S.H.M. with a period of 6.28 s isA. parabolaB. straight lineC. ellipseD. circle |
| Answer» Correct Answer - D | |
| 947. |
If displacement time equation of an S.H.M. is`x= sin . (pit)/(6)` , then the moment of times at which peak value of acceleration is attained areA. 2,4,6,…....sB. 1,3,5 ….....sC. 3,9,15 ….....sD. 1,2,3…......s |
| Answer» Correct Answer - 3 | |
| 948. |
Graph between velocity and displacement of a particle, executing S.H.M. isA. parabolaB. straight lineC. ellipseD. circle |
| Answer» Correct Answer - C | |
| 949. |
A system executing `SHM` must possessesA. inertia onlyB. restoring force onlyC. both restoring force and inertiaD. only extermal force. |
| Answer» Correct Answer - C | |
| 950. |
If a particle is executing S.H.M. then the graph between its acceleration and velocity is , in generalA. An ellipseB. A circleC. A parabolaD. A hyperbola |
| Answer» Correct Answer - 1 | |