Oscillations practice · Mixed difficulty

The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is $16\,\,cm{s^{ - 1}}$ The distance of the particle from the mean position at which the speed of the particle become $8\sqrt 3 cm{s^{ - 1}},$ will be

Which of the following quantities is always negative in SHM?
Here, $\mathbf{s}$ is displacement from mean position.

The circular motion of a particle with constant speed is

Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x-axis. Their mean position is separated by distance ${x_0}$ $({x_0} > A)$. If the maximum separation between them is $\left( {{x_0} + A} \right)$, the phase difference between their motions is

A uniform rod of length $l$ is suspended from any point $P$. It is then made to undergo small oscillations. If $P$ is the

The path difference between the two waves: ${y_1} = {a_1}\,\sin \left( {\omega t - \frac{{2\pi x}}{\lambda }} \right)\,$ and ${y_2} = {a_2}\,\sin \left( {\omega t - \frac{{2\pi x}}{\lambda } + \phi } \right)\,$ will be

Two particles are executing simple harmonic motion. At an instant of time $t$, their displacements are $y_1=A \cos (\omega t)$ and $y_2=A \sin (\omega t)$
Then, the phase difference between $y_1$ and $y_2$ is

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of ${10^{ - 2}}m$. The relative change in the angular frequency of the pendulum is best given by

A particle is attached to a vertical spring and is pulled down a distance $0.04 \mathrm{~m}$ below its equilibrium position and is released from rest. The initial upward acceleration of the particle is $0.30 \mathrm{~ms}^{-2}$. The period of the oscillation is

A block whose mass is $2 \mathrm{~kg}$ is fastened to a spring whose spring constant is $100 \mathrm{Nm}^{-1}$. It is pulled to a distance of $0.1 \mathrm{~m}$ from over a frictionless surface and is released at $t=0$. Calculate the kinetic energy of the block when it is $0.05 \mathrm{~m}$ away from its mean position.

The time period of oscillation of a simple pendulum of length $L$ suspended from the roof a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by:

Statement I When a wave goes from one medium to other, average power transmitted by the wave may change.

Statement II Due to a change in the medium, amplitude, speed, wavelength and frequency of the wave may change.

Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion

Assertion (A) : If the elevator goes down with constant acceleration, then frequency of oscillation of spring block system in elevator will be independent of acceleration of elevator.
Reason (R) : If the given elevator moves with uniform velocity, then the frequency of oscillation will change.

Statement-I: If a pendulum clock is taken to mountain top, it gains time.
Statement-II: Value of acceleration due to gravity is less at heights.

Assertion: In simple harmonic motion, the motion is to and fro and periodic
Reason: Velocity of the particle in simple harmonic motion $(v)=\omega \sqrt{{k}^2-{x}^2}$ (where ${x}$ is the displacement and ${k}$ is spring factor).

In S.H.M. which one of the following quantities has constant ratio with acceleration:

A uniform spring whose unstretched length is $l$ has a force constant $k$. The spring is cut into two pieces of unstretched lengths $l_1$ and $l_2$, where $l_1=n l_2$ and $n$ is being an integer. Now, a mass $m$ is made to oscillate with first spring. The time period of its oscillation would be

The equation of a travelling wave is $y=15 \sin ^2(3 x-4 t),$ where $x, y$ are in meter and $t$ is in second. The amplitude of oscillation of particle is:

Sine and cosine functions are:

The displacement of a particle in SHM various according to the relation $x = 4\left( {\mathrm{cos\pi }t + \mathrm{sin\pi }t} \right)$ The amplitude of the particle is

The angular frequency of a simple pendulum is $\omega \,rad/s$. Now the length is made one fourth of the original length , the angular frequency becomes

Displacement between maximum potential energy position and maximum kinetic energy position for a particle executing S.H.M. is

The equation of SHM is given by
$x = 3\sin 20\pi t + 4\cos 20\pi t$
Where x is in cm and t is I second. The amplitude is

Two simple harmonic motion are represented by ${y_1} = 5\left( {\sin 2\pi t + \sqrt 3 \cos 2\pi t} \right)$ ${y_2} = 5\sin \left( {2\pi t + \frac{\pi }{4}} \right)$ The ratio of the amplitudes of two SHM’s is

To show that a simple pendulum executes simple harmonic motion, it is necessary to assume that

If the equation $y_1=A \sin \omega t$ and $y_2=\frac{A}{2} \sin \omega t+\frac{A}{2} \cos \omega t$ represents SHM, then the ratio of the amplitudes of the two motions is

A particle moves with simple harmonic motion in a straight line. In first $\tau$ sec, after starting from rest it travels a distance $a$ and in next $\tau$ sec, it travels $2a$, in same direction, then

A particle in UALM is moving with acceleration $2 \mathrm{~m/s}^2$ and initial velocity 20 m/s. Find time of journey if displacement is 96 m

$x(t)=A \cos (\omega t+\phi)$ is the equation of simple harmonic motion. In this equation ' $\phi$ ' is called