Oscillations practice · Mixed difficulty

The amplitude of SHM $\;y = 2\left( {\sin 5\,\,\mathrm{\pi }t + \sqrt 2 \mathrm{cos}\,\,\mathrm{\pi }t} \right)$ is

Two pendulums of length 1 m and 16 m start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is T. the minimum time after which the two threads of the pendulum will be one behind the other is?

A second pendulum is moved to moon, where acceleration due to gravity is $1 / 6$ times that of the earth, the length of the second pendulum on moon would be

The position vector of a particle $\overrightarrow R$ as a function of time is given by:
$\mathop R\limits^ - \, = \,4\,\sin \left( {2\pi t} \right)\mathrm{\hat I}\, + \,4\,\cos \,\left( {2\pi t} \right)\mathrm{\hat J}$
Where R is in meters, t is in seconds and $\mathrm{\hat I}$ and $\mathrm{\hat J}$ denote unit vectors along x- and y-directions, respectively. Which
One of the following statements is wrong for the motion of particle?

Identify the correct statement.

If $Y=A \sin (2 \pi m t+\phi)$ represents SHM of a particle then its phase at time $t$ is

Two simple pendulums of length $1 \mathrm{~m}$ and $1.21 \mathrm{~m}$ are started oscillating from some position. Find the minimum time after which they again start from same position.

If a simple pendulum of length l has maximum angular displacement θ, then the maximum kinetic energy of bob of mass m is

The amplitude of a damped oscillator becomes half in one minute. The amplitude after 3 minute will be $\frac{1}{X}$ times the original, where $X$ is

The displacement x(in metre) of a particle in simple harmonic motion is related to time t(in second) as $x = 0.01\cos \left( {\pi t + \frac{\pi }{4}} \right)$ The frequency of the motion will be

A displacement wave is represented by $\mathrm{y}=0.25 \times 10^{-3} \sin (500 t-0.025 x)$, where $y$, $t$ and $x$ are in $\mathrm{cm}$, sec and metre respectively. Deduce the amplitude of particle acceleration.

How does time period of a pendulum very with length?

I. Average kinetic energy in one oscillation during SHM of a body is $\frac{1}{4} m \omega^2 A^2$.

II. Maximum kinetic energy is $\frac{1}{2} m \omega^2 A^2$.

Which of the following statement(s) is/are correct?

The mass of a bob, suspended in a simple pendulum, is halved from the initial mass, its time period will:

When a particle executes Simple Harmonic Motion, the nature of graph of velocity as a function of displacement will be :

The total energy of a particle, executing simple harmonic motion is Where x is the displacement from the mean position.

A particle moves in x-y plane according to rule x=a sin⁡ωt and y=a cos⁡ωt. The particle follows

A particle free to move along the $x$ -axis has potential energy given as $U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right]\,\,\,\,\,\,\left( {\mathrm{for} - \infty \le + \infty } \right)$ Where k is a positive constant of appropriate dimensions. Then

The total energy of a simple harmonic oscillator is proportional to

The phenomenon that occurs when the
frequency of forced vibrations on an object
matches the natural frequency of that object,
and produces a dramatic increase in amplitude
is called

A particle of mass 200 g executes SHM. The restoring force is provided by a spring of force constant 80 N/m. The time period of oscillation is

In SHM, phase difference between velocity and acceleration is

If the displacement equation of a particle be represented by $y = A\sin PT + B\cos PT,$ the particle executes

A body oscillates in SHM according to the equation (in SI unit), $x=5 \cos \left(2 \pi t+\frac{\pi}{4}\right)$. Its instantaneous displacement at $t=1 \mathrm{~s}$ is

A particle is executing simple harmonic motion with a period of $T$ seconds and amplitude a metre. The shortest time it takes to reach a point $\frac{a}{{\sqrt 2 }}m$ from its mean position in seconds is

The velocities of a particle in SHM at positions $x_1$ and $x_2$ are $v_1$ and $v_2,$ respectively, its time period will be

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :

Statement-1: A particle is moving along $x$-axis. The resultant force $F$ acting on it is given by $F=-a x-b$ where $a$ and $b$ are both positive constants. The motion of this particle is not S.H.M.
Statement-2: In S.H.M. resultant force must be proportional to the displacement from mean position.

A solid cube floats in water half immersed and has small vertical oscillations of time period $\frac{\pi}{5} \mathrm{~s}$. Its mass (in kg) is (Take, $g=10 \mathrm{~ms}^{-2}$ )

The time period of a simple pendulum on earth is $T.$ If it is taken on the moon, and made to oscillate, the period of vibration will be: