Periodic Motion & SHM Basics

The differential equation of a particle performing S.H.M. is $\frac{d^2x}{dt^2} + 64x = 0$, where $x$ is in meter. The period of oscillation of the particle is (in second)

The period of S.H.M. of a particle is $12$s. The phase difference between the positions at $t = 3$s and $t = 4$s will be

The minimum phase difference between two simple harmonic motions is

A mass attached to a spring performs S.H.M. whose displacement is $x = 3 \times 10^{-3}\cos 2\pi t$ m. The time taken to obtain maximum speed for the first time is (in s)

A ball released from height $(h)$ which makes perfectly elastic collision with ground. The frequency of periodic vibratory motion is ($g$ = acceleration due to gravity)

Match List - I with List - II.

Choose the correct answer from the options given below:

Two simple harmonic motions are represented as $y_1 = 10\sin \omega t$ and $y_2 = 5\sin \omega t + 5\cos \omega t$, both in cm. The ratio of the amplitudes of $y_2$ ($A_2$) to that of $y_1$ ($A_1$ ), is

The displacement of a particle varies with time according to the relation

$x = a\sin \omega t + b\cos \omega t$

The displacement of a particle performing linear S.H.M. is given by $y = A\cos[\pi(t + \phi)]$. If at $t = 0$, the displacement is $y = 2$ cm and velocity is $2\pi$ cm/s, the value of amplitude A in cm is

A particle oscillates simple harmonically along a straight line with period $8$ s and amplitude $8\sqrt{2}$ m. It starts from mean position then the ratio of the distances travelled by it in the $2^{nd}$ second and the first second of its motion is $\left(\sin 45^\circ = \frac{1}{\sqrt{2}}\right)\ (\sin 90^\circ = 1)$

A particle performing S.H.M., has a period of $12$ s. The particle starts from a mean position. The ratio of distances travelled by the particle in the first and second second is $\left\{\cos(\pi/6) = \frac{\sqrt{3}}{2}\right\}$

A particle executes a simple harmonic motion with a periodic time $8$ second. At time $t = 0$, it is at a mean position. The ratio of the distance traveled by a particle in the 2nd and that in the 1st second of its motion is

($\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}},\ \sin 90^\circ = \cos 0^\circ = 1$)

A particle executes simple harmonic motion between $x = -A$ and $x = +A$. If time taken by particle to go from $x = 0$ to $\frac{A}{2}$ is $2$s, then time taken by particle in going from $x = \frac{A}{2}$ to A is

A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec. The time required by it to move from $x = A$ to $x = \frac{A}{\sqrt{2}}$ is ______ sec.

The equation of motion of a particle is given by $x = a\sin(50t + \pi/3)$ cm. The particle will come to rest at time $t_1$ and it will have zero acceleration at time $t_2$. The $t_1$ and $t_2$ respectively are ________.

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Knowing initial position $x_0$ and initial momentum $p_0$ is enough to determine the position and momentum at any time $t$ for a simple harmonic motion with a given angular frequency $\omega$.

Reason (R): The amplitude and phase can be expressed in terms of $x_0$ and $p_0$.

In the light of the above statements, choose the correct answer from the options given below:

The period of S. H.M. of a particle is 16 second. The phase difference between the positions at $t = 2\text{ s}$ and $t = 4\text{ s}$ will be

The displacement of particle in S.H.M. is $x = A\cos(\omega t + \pi/6)$. Its speed will be maximum at time $(\sin 90^\circ = 1)$

A mass attached to a spring performs S.H.M. whose displacement is $x = 3 \times 10^{-3}\cos 2\pi t$ metre. The time taken to obtain maximum speed for the first time is

The motion of the particle is given by the equation $x = A\sin\omega t + B\cos\omega t$.

The motion of the particle is

Two particles ' $A$ ' and ' $B$ ' execute SHMs of periods ' $T$ ' and $\frac{3T}{2}$. If they start from the mean position then the phase difference between them, when the particle ' $A$ ' completes two oscillations will be

A particle performing linear S.H.M. has period 8 seconds. At time $t = 0$, it is in the mean position. The ratio of the distances travelled by the particle in the $1$st and $2$nd second is $\left(\cos 45^\circ = 1/\sqrt{2}\right)$

A particle oscillates in straight line simple harmonically with period 8 second and amplitude $4\sqrt{2}\text{ m}$. Particle starts from mean position. The ratio of the distance travelled by it in $1$st second of its motion to that in $2$nd second is $\left(\sin 45^\circ = 1/\sqrt{2}, \sin\frac{\pi}{2} = 1\right)$

A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then $\frac{D}{d}$ is

The motion of a particle is described by the equation $a = -bx$ where ' $a$ ' is the acceleration, x is the displacement from the equilibrium position and b is a constant. The periodic time will be

When a particle in linear S.H.M. completes two oscillations, its phase increases by

If $x = 5\sin\left(\pi t + \frac{\pi}{3}\right)\text{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

A particle of mass $0.50\text{ kg}$ executes simple harmonic motion under force $F = -50(\text{N m}^{-1})x$. The time period of oscillation is $\frac{x}{35}\text{ s}$. The value of $x$ is _________.

(Given $\pi = \frac{22}{7}$)

The displacement of a particle performing S.H.M. is given by $Y = A\cos[\pi(t + \phi)]$. If at $t = 0$, the displacement is $y = 2\text{ cm}$ and velocity is $2\pi\text{ cm/s}$, the value of amplitude $A$ in cm is

A simple harmonic oscillator has an amplitude $A$ and time period $6\pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x =$ A to $x = \frac{\sqrt{3}}{2}$ A will be $\frac{\pi}{x}\text{ s}$, where $x =$ _________.

In a linear Simple Harmonic Motion (SHM)

(A) Restoring force is directly proportional to the displacement.

(B) The acceleration and displacement are opposite in direction.

(C) The velocity is maximum at mean position.

(D) The acceleration is minimum at extreme points.

Choose the correct answer from the options given below:

For particle P revolving round the centre O with radius of circular path $r$ and angular velocity $\omega$, as shown in below figure, the projection of OP on the $x$-axis at time $t$ is

For a periodic motion represented by the equation

$y = \sin \omega t + \cos \omega t$

the amplitude of the motion is

A body of mass $0.04\text{ kg}$ executes simple harmonic motion (SHM) about $x = 0$ under the influence of force $F$ as shown in graph. The period of oscillation is

Two S.H.Ms. are represented by equations $y_1 = 0.1\sin\left(100\pi t + \frac{\pi}{3}\right)$ and $y_2 = 0.1\cos(100\pi t)$ The phase difference between the speeds of the two particles is

A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is $x = \frac{A}{2}$ and it moves along positive x-axis. The displacement of particle in time t is $x = A\sin(\omega t + \delta)$, then the value of $\delta$ will be

A particle executes simple harmonic motion between $x = -A$ and $x = +A$. If time taken by particle to go from $x = 0$ to $\frac{A}{2}$ is 2 s; then time taken by particle in going from $x = \frac{A}{2}$ to A is

A simple harmonic progressive wave is represented by $y = A\sin(100\pi t + 3x)$. The distance between two points on the wave at a phase difference of $\frac{\pi}{3}$ radian is

Identify the function which represents a non-periodic motion.

The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is:

The potential energy of a particle of mass $4\text{ kg}$ in motion along the x-axis is given by $U = 4(1 - \cos 4x)$ J. The time period of the particle for small oscillation $(\sin\theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right)\text{ s}$. The value of $K$ is _________.

The time taken by a particle executing simple harmonic motion of period '$T$', to move from the mean position to half the maximum displacement is

A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ___________ s.

In a medium, the phase difference between two particles separated by a distance '$x$' is $\left(\frac{\pi}{5}\right)$c. If the frequency of the oscillation of particles is $25\text{ Hz}$ and the velocity of propagation of the waves is $75\text{ m/s}$, then the value of $x$ is

The function of time representing a simple harmonic motion with a period of $\frac{\pi}{\omega}$ is:

Y = A sin($\omega$t + $\phi$0) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is $Y = \frac{A}{2}$ and it is moving along negative x-direction. Then the initial phase angle $\phi$0 will be:

A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a}$s. The value of 'a' to the nearest integer is _________.

Two particles $P$ and $Q$ performs S.H.M. of same amplitude and frequency along the same straight line. At a particular instant, maximum distance between two particles is $\sqrt{2}$ a. The initial phase difference between them is

$\left[\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\right]$

The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers 30$^\circ$ in 0.1 s. The perpendicular projection 'P' from 'A' on the diameter MN represents the simple harmonic motion of 'P'. The restoration force per unit mass when P touches M will be:

Two simple harmonic motion, are represented by the equations $y_1 = 10\sin\left(3\pi t + \frac{\pi}{3}\right) \quad y_2 = 5(\sin 3\pi t + \sqrt{3}\cos 3\pi t)$ Ratio of amplitude of y1 to y2 = x: 1. The value of x is ______________.

Two simple harmonic motions are represented by the equations

$x_1 = 5\sin\left(2\pi t + \frac{\pi}{4}\right)$ and $x_2 = 5\sqrt{2}(\sin 2\pi t + \cos 2\pi t)$. The amplitude of second motion is ................ times the amplitude in first motion.

A particle executes simple harmonic motion represented by displacement function as x(t) = A sin($\omega$t + $\phi$)

If the position and velocity of the particle at t = 0 s are 2 cm and 2$\omega$ cm s${}^{-1}$ respectively, then its amplitude is $x\sqrt{2}$ cm where the value of x is _________________.

Time period of a simple pendulum is T. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha}{\beta}T$. The value of $\alpha$ is _________.

'$n$' waves are produced on a string in 1 second. When the radius of the string is doubled, keeping tension same, the number of waves produced in 1 second for the same harmonic will be

Identify the function which represents a periodic motion.

A particle performs simple harmonic motion with period of 3 s. The time taken by it to cover a distance equal to half the amplitude from mean position is [$\sin 30^\circ = 0.5$]

The radius of circle, the period of revolution, initial position and sense of revolution are indicated is the figure.

y- projection of the radius vector of rotating particle P is:

A simple harmonic motion is represented by:

y = 5 (sin 3 $\pi$ t + $\sqrt{3}$ cos 3 $\pi$t) cm

The amplitude and time period of the motion are:

A particle executes the simple harmonic motion with an amplitude ' $A$ '. The distance travelled by it in one periodic time is

The distance covered by a particle undergoing SHM in one time period is (amplitude = $A$),

Average velocity of a particle executing SHM in one complete vibration is:

The displacement of a particle executing simple harmonic motion is given by

y = A0 + A sin$\omega$t + B cos$\omega$t.

Then the amplitude of its oscillation is given by:

A particle executes simple harmonic motion and is located at x = a, b and c at times t0, 2t0 and 3t0 respectively. The frequency of the oscillation is:

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t = 0 one particle has displacement A while the other one has displacement $\frac{-A}{2}$ and they are moving towards each other. If they cross each other at time t, then t is:

Which of the following expressions corresponds to simple harmonic motion along a straight line, where x is the displacement and a, b, c are positive constants?

A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

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

Out of the following functions representing motion of a particle which represents SHM (1) y = sin$\omega$t $-$ cos$\omega$t

(2) y = sin3$\omega$t

(3) y = 5cos $\left(\frac{3\pi}{4} - 3\omega t\right)$

(4) y = 1 + $\omega$t + $\omega$2t2

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean position is separated by distance $X_0\ (X_0 > A)$. If the maximum separation between them is $(X_0 + A),$ the phase difference between their motion is:

The displacement of a particle along the x-axis is given by x = asin2$\omega$t. The motion of the particle corresponds to

Which one of the following equations of motion represents simple harmonic motion? where k, k0, k1 and a are all positive.

A point mass oscillates along the $x$-axis according to the law $x = x_0 \cos(\omega t - \pi/4).$ If the acceleration of the particle is written as $a = A\cos(\omega t + \delta),$ then

A particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

The displacement of an object attached to a spring and executing simple harmonic motion is given by $x = 2 \times 10^{-2}\cos \pi t$ metre. The time at which the maximum speed first occurs is

The circular motion of a particle with constant speed is

If a simple harmonic motion is represented by $\frac{d^2x}{dt^2} + \alpha x = 0.$ its time period is

The function $\sin^2(\omega t)$ represents

Two simple harmonic motions are represented by the equations $y_1 = 0.1\sin\left(100\pi t + \frac{\pi}{3}\right)$ and $y_2 = 0.1\cos \pi t.$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is

The displacement of particle varies according to the relation

$x = 4(\cos \pi t + \sin \pi t).$ The amplitude of the particle is

A particle executes S.H.M. along $x$-axis. The force acting on it is given by

A simple harmonic oscillator has an amplitude $A$ and time period $T$. The time required by it to travel from $x = A$ to $x = A/2$ is

A particle executes S.H.M. along $x$-axis. The force acting on it is given by