Oscillations practice · Mixed difficulty

A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particle after 2T period from its initial position is

The kinetic energy of a particle executing S.H.M. is 16 J when it is at its mean position. If the mass of the particle is 0.32 kg, then what is the maximum velocity of the particle

The potential energy of a particle in SHM is:

${T_0}$ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to $\frac{1}{{16}}$ times of its initial value, then the modified time period is

A simple pendulum consists of a mass attached to a light string $l.$ if the system is oscillating through small angles which of the following is true?

The relation between acceleration and displacement of four particles are given below. Which one of the particles is executing simple harmonic motion?

Force acting on a body moving along $x$-axis is given as $F=-k x^3$, where $x$ is the displacement of the body from its mean position. The motion of body is:

Statement-1: Time period of spring pendulum is the same whether in an accelerated or in an inertial frame of reference.
Statement-2: Mass of the bob of spring pendulum and spring constant of the spring are independent of the acceleration of the frame of reference.

The motion of a simple pendulum executing S.H.M. is represented by the following equation.
$y = A\sin \left( {\pi t + \phi } \right)$, where time is measured in second. The length of pendulum is

In the differential equation of linear simple harmonic motion, $\frac{d^2 x}{d t^2}+\omega^2 x=0$, the term $\omega^2$ represents

The velocity vector $v$ and displacement vector $x$ of a particle executing SHM are related as $\frac{v d v}{d x}=-\omega^2 x$ with the initial condition $v=v_0$ at $x=0$. The velocity $v$ when displacement is $x$, is

A simple wave motion is represented by $y=5(\sin \,4 \,\pi\, t+\sqrt{3} \cos \,4 \,\pi \,t)$. Its amplitude is

What is the effect on the time period of a simple pendulum if the mass of the bob is doubled

Starting from $y = A$ $\sin \,\,\omega t\mathrm{ }or\mathrm{ }y = A\,\,\cos \,\,\omega t$

The physical quantity conserved in simple
harmonic motion is

A mass $m$ attached to a spring of spring constant $k$ is stretched a distance $x_0$ from its equilibrium position and released with no initial velocity. The maximum speed attained by the mass in its subsequent motion and the time at which this speed would be attained are respectively,

Assertion: Amplitude of a forced vibration can never reach infinity.
Reason: The driving frequency cannot be increased beyond a certain limit.

Consider the following statements : The total energy of a particle executing simple harmonic motion depends on its I. Amplitude II. Period III. Displacement Of these statements

For Simple Harmonic Oscillator, the potential energy is equal to kinetic energy

A particle executing SHM while moving from one extremity is found at distances $x_1, x_2$ and $x_3$ from the centre at the end of three successive seconds. The time period of oscillation is

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

A particle executes $SHM$ with an amplitude of $5\,cm$. The percentage of its energy which is potential energy at a displacement of $1\,cm$ from the mean position is

The amplitude of oscillations of a simple harmonic oscillator of mass 100 gram having an angular frequency of $10\,\,rad\,{s^{ - 1}}$ with a total mechanical energy of 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 simple harmonic motion is represented by $x(t)=\sin ^2 \omega t-2 \cos ^2 \omega t$. The angular frequency of oscillation is given by

The period of oscillation of a simple pendulum is given by $T=2 \pi \sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity. The length is measured using a meter scale which has 2000 divisions. If the measured value $L$ is $50 \mathrm{~cm}$, the accuracy in the determination of $g$ is $1.1 \%$ and the time taken for 100 oscillations is 100 seconds, what should be the resolution of the clock (in milliseconds)?

The potential energy of a particle of mass $1 \mathrm{~kg}$ in motion along the $x$-axis is given by $U=4(1-\cos 2 x) \mathrm{J}$, where $x$ is in metres. The period of small oscillations (in second) is

The motion of a particle in SHM is of

In order that the resultant path on superimposing two mutually perpendicular SHM be a circle, the conditions are that

Time period of a simple pendulum in a stationary lift is $'T'$ If the lift accelerates with $\frac{g}{6}$ vertically upwards then the time period will be
(Where $g$ = acceleration due to gravity)