Physics - Simple Harmonic Motion
Exam Duration: 45 Mins Total Questions : 30
Amplitude is a
- (a)
scalar quantity
- (b)
vector quantity
- (c)
cannot be negative
- (d)
Both (a) and (c) are correct
If the maximum velocity in SHM is vmax. Then, the average velocity during motion from one extreme point to the other extreme point will be
- (a)
\(\frac{4}{\pi}v_{max}\)
- (b)
\(\frac{\pi}{4}v_{max}\)
- (c)
\(\frac{2}{\pi}v_{max}\)
- (d)
\(\frac{\pi}{2}v_{max}\)
A body doing SHM with amplitude 1 cm and frequency 60 Hz. The maximum acceleration will be
- (a)
\(200 \quad\pi^2m/s\)
- (b)
\(400 \quad\pi^2m/s^2\)
- (c)
\(244 \quad\pi^2m/s^2\)
- (d)
\(144 \quad\pi^2m/s^2\)
A body executing SHM of time period 4 s. Time taken by it to move from mean position to half of amplitude starting from the mean position is
- (a)
4 s
- (b)
\(\frac{1}{\sqrt{3}}s\)
- (c)
\(\frac{1}{3}s\)
- (d)
\(\frac{2}{3}s\)
A particle is acted simultaneously by mutually perpendicular forces then nature of simple harmonic motion of the particle will be
- (a)
an ellipse
- (b)
a parabola
- (c)
a circle
- (d)
a straight line
The ratio of amplitudes of following SHM is x1=\(A \ sin \ \omega t\) and
- (a)
\(\sqrt {2}\)
- (b)
\(\frac{1}{\sqrt{2}}\)
- (c)
1
- (d)
2
A particle executing SHM with frequency v. The frequency with which kinetic energy oscillate is
- (a)
4v
- (b)
v
- (c)
v/2
- (d)
2v
If <E> and <V> denotes the average kinetic and average potential energies respectively of mass describing a simple harmonic motion over one period, then the correct relation is
- (a)
<E> = <V>
- (b)
<E> = 2<V>
- (c)
<E> = -2<V>
- (d)
<E> = -<V>
Two SHM x1=A sin ωt and x2=A cos ωt are superimposed on a particle having mass m. Total mechanical energy of particle is
- (a)
zero
- (b)
\(\frac{1}{4}\ m \ \omega^2A^2\)
- (c)
\(\frac{1}{2}\ m \ \omega^2A^2\)
- (d)
\(m \ \omega^2A^2\)
In string of simple pendulum, tension in the string is
- (a)
maximum at mean position
- (b)
minimum at mean position
- (c)
zero
- (d)
constant
A lift is ascending with acceleration g/3. What will be the time period of a simple pendulum suspended from its ceiling, if its time period in stationary lift is T?
- (a)
\(T/2\)
- (b)
\(T/4\)
- (c)
\(\sqrt{3}T/4\)
- (d)
\(\sqrt{3}T/2\)
A damped oscillator of frequency v1 is driven by an external periodic force of frequency v2. when steady state is reached, frequency of oscillator will be
- (a)
\(v_{2}\)
- (b)
\(v_{1}\)
- (c)
\(\frac{v_1+v_2}{2}\)
- (d)
\(\sqrt{v_1+v_2}\)
The time period of a simple pendulum is T. when length is increased by 10 cm, its period is T1. When length is decreased by 10 cm, its period is T2. Then relation between T, T1 and T2 is
- (a)
\(2T^2 = T_1^2 - T_2^2 \)
- (b)
\(2T^2 = T_1^2 + T_2^2 \)
- (c)
\(\frac{2}{T^2}=\frac{1}{T_1^2}-\frac{1}{T_2^2}\)
- (d)
\(\frac{2}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}\)
A point mass m is suspended at the end of a massless wire of length L and cross-section area A. If Y is the Young's modulus for wire, then the frequency of oscillation for SHM along vertical line is
- (a)
\(\frac{1}{2\pi}\sqrt{\frac{mL}{YA}}\)
- (b)
\(\frac{1}{2\pi}\sqrt{\frac{YA}{mL}}\)
- (c)
\(\frac{1}{\pi}\sqrt{\frac{YA}{mL}}\)
- (d)
\({\pi}\sqrt{\frac{YA}{mL}}\)
Match the physical parameters given in column I with their formula given in column II and select the correct options from the choices given below
Column I | Column II |
A. Time period of spring B. Angular acceleration C. Time period of simple pendulum |
1. \(\frac{d^2\theta}{dt^2}\alpha-\theta\) 2. \(T=2\pi\sqrt{l/g}\) 3. \(T=2\pi\sqrt{m/K}\) |
- (a)
A B C 1 2 3 - (b)
A B C 3 1 2 - (c)
A B C 3 2 1 - (d)
A B C 1 3 2
A particle moves with simple harmonic motion in a straight line. In first \(\tau\) second, after starting from rest it travels a distance a, and in next \(\tau\) second it travels 2a, in same direction, then
- (a)
amplitude of motion is 4a
- (b)
time period of oscillations is \(6\tau\)
- (c)
amplitude of motion is 3a
- (d)
time period of oscillations is \(8\tau\)
If a simple pendulum has significant amplitude (upto a factor of 1/e of original) only in the period between t=0 sec to t=\(\tau\) sec, then \(\tau\) may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation proportional to its velocity with b as the constant of proportionality, the average life time of the pendulum is ( assuming damping is small) in seconds
- (a)
\(\frac{0.0693}{b}\)
- (b)
\(b\)
- (c)
\(\frac{1}{b}\)
- (d)
\(\frac{2}{b}\)
Two particles are executing simple harmonic motion of the same amplitude A and frequency \(\omega\) along the X-axis. Their mean positions are separated by distance X0(X>A). If the maximum separation between them is (X0 + A), the phase difference between their motion is
- (a)
\(\frac{\pi}{6}\)
- (b)
\(\frac{\pi}{2}\)
- (c)
\(\frac{\pi}{3}\)
- (d)
\(\frac{\pi}{4}\)
A mass M, attached to a horizontal spring, executes SHM with amplitude A1. When the mass M passes through its mean position, then a smaller mass m is placed over it and both of them move together with amplitude A2. The ratio of (A1/A2) is
- (a)
\(\frac{M+m}{M}\)
- (b)
\(\left(\frac{M} {M+m}\right)^{1/2}\)
- (c)
\(\left(\frac{M+m}{M} \right)^{1/2}\)
- (d)
\(\frac{M}{M+m}\)
If x,v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then which of the following does not change with time?
- (a)
\(a^2T^2+4\pi^2v^2\)
- (b)
\(aT/x\)
- (c)
\(aT+2 \pi v\)
- (d)
\({aT}/{v}\)
A point mass oscillates along the X-axis according to the law \(x=x_0cos(\omega t-\frac{\pi}{4})\).
If the acceleration of the particle is written as \(a=Acos(\omega t-\delta)\), then
- (a)
\(A=x_0, \delta=-\frac{\pi}{4}\)
- (b)
\(A=x_0 \omega^2, \delta=\frac{\pi}{4}\)
- (c)
\(A=x_0 \omega^2, \delta=-\frac{\pi}{4}\)
- (d)
\(A=x_0 \omega^2, \delta=\frac{3\pi}{4}\)
The displacement of an object attached to a spring and executing simple harmonic motion is given by \(x=2 \times10^{-2} \ cos \ \pi t\) metre. The time at which the maximum speed first occurs
- (a)
0.5 s
- (b)
0.75 s
- (c)
0.125 s
- (d)
0.25 s