A body oscillates with SHM according to the equation \\(x = 5.0 cos\left( {2\pi t + \pi } \right)\\). At time \\(t=1.5 s\\), its displacement, speed and acceleration respectively is:

Given,\\(x=5cos(2\pi t+\pi)\\)\\(t=1.5 sec\\)Displacement, \\(x=5cos(2\pi \times 1.5+\pi)=5cos(4\pi)\\)\\(x=5m\\). . . . . .(1)Velocity, \\(v=\dfrac{dx}{dt}\\)\\(v=-10\pi sin(2\pi t+\pi)\\)\\(v=-10\pi sin(2\pi \times 1.5+\pi)\\)\\(v=-10\pi sin(4\pi)\\)\\(v=0m/s\\). . . . .(2)Acceleration, \\(a=\dfrac{dv}{dt}\\)\\(a=-20\pi ^2cos(2\pi t +\pi)\\)\\(a=-20\pi^2cos(2\pi \times 1.5+\pi)\\)\\(a=-20\pi^2cos(4\pi)\\)\\(a=-20\pi^2\\)The correct option is B.

A particle is executing SHM along a straight line. Its velocities at distances \\(x_1\\) and \\(x_2\\) from the mean position are \\(V_1\\) and \\(V_2\\), respectively. Its time period is:

Using the expressions for velocity for SHM,\\(v_1 = \omega \sqrt{A^2 -x_1^2}\\)\\(v_2 = \omega \sqrt{A^2 -x_2^2}\\)Squaring both,\\(v_1 ^2 = \omega ^2 ( A^2 -x_1 ^2)\\) ....(3)\\(v_2 ^2 = \omega ^2 ( A^2 -x_2 ^2)\\) ....(4)Subtracting (4) from (3), \\(v_1 ^2 - v_2 ^2 = \omega ^2 ( x_2 ^2 - x_1 ^2) \\)\\( \Rightarrow 4 \pi ^2 ( x_2 ^2 - x_1 ^2) = T^2 (v_1 ^2 - v_2 ^2 )\\)\\( \Rightarrow T = 2\pi \sqrt{\dfrac{ x_2 ^2 - x_1 ^2}{v_1 ^2 - v_2 ^2}}\\) where we have used \\(\omega = \dfrac{2\pi}{T}\\)

A particle executes linear simple harmonic motion with an amplitude of \\(3\ cm\\). When the particle is at \\(2\ cm\\) from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Velocity of particle executing S.H.M \\(v=\omega \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \\) and acceleration, \\(a=-{ \omega }^{ 2 }x\\) \\(x\\)= displacement at any time interval\\(a\\) = amplitude\\(=3cm\\)\\(\omega\\)= angular frequency, \\(=\cfrac { 2\pi }{ T } \\)\\(T\\)=Time periodNow, at \\(x=2cm\\) \\(v=|a|\\) \\(\Rightarrow { \omega }^{ 2 }\times 2=\omega \sqrt { { 3 }^{ 2 }-{ 2 }^{ 2 } } \\)\\(\Rightarrow { \omega }=\cfrac { \sqrt { 5 } }{ 2 } \\)\\(\Rightarrow \cfrac { 2\pi }{ T } =\cfrac { \sqrt { 5 } }{ 2 } \\)\\(\Rightarrow T=\cfrac { 4\pi }{ \sqrt { 5 } } \\)

A simple pendulum performs simple harmonic motion about \\(x = 0\\) with an amplitude \\('a'\\) and time period \\('T'\\). The speed of the pendulum at \\(x = \dfrac{a}{2}\\) will be:

As simple pendulum performs simple harmonic motion.\\(\therefore velocity, v = \omega \sqrt {a^{2} - x^{2}}\\)At, \\(x = \dfrac {a}{2}\\)\\(v = \dfrac {2\pi}{T} \sqrt {a^{2} -\left (\dfrac {a}{2}\right )^{2}} = \dfrac {2\pi}{T} \dfrac {\sqrt {3a^{2}}}{2} = \dfrac {\pi a\sqrt {3}}{T}\\)

A particle is executing a simple harmonic motion. Its maximum acceleration is \\(\alpha\\) and maximum velocity is \\(\beta\\). Then, its time period of vibration will be:

For SHM, maximum acceleration, \\(a_{max}=\omega^2 a=\alpha\\) where a is the amplitude of SHM. and maximum velocity, \\(v_{max}=\omega a=\beta\\) so, \\(\dfrac{\alpha}{\beta}=\dfrac{\omega^2 a}{\omega a}=\omega\\) or \\(\dfrac{\alpha}{\beta}=\dfrac{2\pi}{ T}\\) or \\(T=\dfrac{2\pi \beta}{\alpha}\\)

If the displacement (x) and velocity (v) of a particle executing S.H.M. are related through the expression \\(4v^2= 25 - x^2\\), then its time period is :

Velocity in SHM is given as\\(V^{2}=\omega^{2}(A^{2}-x^{2})\\)Thus\\(V^{2}=\dfrac{1}{4}(5^{2}-x^{2})\\)\\(\therefore \omega =\dfrac{1}{2}\\)Thus we get the time period as\\(T=\dfrac{2\pi}{\omega}\\)\\(T=4\pi sec\\)

Two simple harmonic motions are represented by the equations \\(y_1=0.1\sin\left(100\pi t+\dfrac{\pi}{3}\right)\\) and \\(y_2=0.1\cos\pi t\\). The phase difference of the velocity of particle \\(1\\) with respect to velocity of particle \\(2\\) at \\(t=0\\) is

\\(y_1 = 0.1sin (100\pi t + \dfrac{\pi}{3})\\)\\(y_2 = 0.1 cos \pi t = 0.1 sin (\dfrac{\pi}{2} + \pi t) = 0.1 sin (\pi t + \dfrac{\pi }{2})\\)Now, for finding velocity of particle , differentiate both equations with respect to time .\\(\dfrac {dy_1}{dt} = v_1 = 0.1 \times 100\pi cos (100\pi t + \dfrac{\pi}{3})\\)similarly for 2nd equation,\\(\dfrac {dy_1}{dt} = v_2 = 0.1 \times \pi cos (\pi t + \dfrac {\pi}{2}) = 0.1 \pi cos (\pi t + \dfrac{\pi}{2})\\)if equation \\(x = Asin (\omega t + \phi)\\) is given then, at \\(t = 0\\) phase of motion is \\(\phi \\)similarly at \\(t = 0\\) phas of 1st particle velocity is \\(\dfrac{\pi}{3}\\)at \\(t= 0 \\) phase of velocity of 2nd particle is \\(\dfrac {\pi}{2}\\)now thw phase difference = phase of 1st particle at \\(t=0\\) - phase of 2nd particle at\\(t = 0\\) \\(= \dfrac{\pi}{3} - \dfrac {\pi}{2} = -\dfrac{\pi}{6}\\)

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance \\(\displaystyle\frac{2A}{3}\\) from equilibrium position. The new amplitude of the motion is.

In SHM, \\(V=\sqrt { { A }^{ 2 }-{ x }^{ 2 } } \\)\\(V=\sqrt { { A }^{ 2 }-\dfrac { 4{ A }^{ 2 } }{ 9 } } =\sqrt { \dfrac { 5{ A }^{ 2 } }{ 9 } } \\)\\(3V=\sqrt { { A' }^{ 2 }-\dfrac { 4{ A }^{ 2 } }{ 9 } } \\)Dividing the above two equations give \\(A'=\dfrac { 7A }{ 3 } \\)

The maximum velocity of a particle, executing simple harmonic motion with an amplitude \\(7\ mm\\), is \\(4.4\ m/s\\). The period of oscillation is

For a SHM, \\(x=A\sin(\omega t +\phi)\\)Velocity, \\(v=\dfrac{dx}{dt}=A\omega \cos(\omega t+\phi)\\)So, \\(v_{max}=A\omega\\)or \\(4.4=7\times 10^{-3}(2\pi/T)\\)or \\(T=9.99\times 10^{-3}=10\times 10^{-3}=0.01 s\\)

The \\(x-t\\) graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at \\(t = \dfrac{4}{3}\ s\\) is

The given motion is represented by\\(\mathrm{x}=1\sin(\pi/4)\mathrm{t}\\)\\(\displaystyle \dfrac{\mathrm{d}^{2}\mathrm{x}}{\mathrm{d}\mathrm{t}^{2}}=\dfrac{-\pi^{2}}{16}\sin(\pi/4)\mathrm{t}\\)At \\(\mathrm{t}=4/3\ \mathrm{s}\mathrm{e}\mathrm{c}\\),\\(\displaystyle \frac{\mathrm{d}^{2}\mathrm{x}}{\mathrm{d}\mathrm{t}^{2}}=-\dfrac{\sqrt{3}}{32}\pi^{2}\ \mathrm{c}\mathrm{m}/\mathrm{s}^{2}\\)

Velocity and Acceleration in SHM | Formulas, Definition, Examples

formula

Velocity in SHM as a function of time

General equation of SHM for displacement in a simple harmonic motion is:

x = A s i n w t

By definition,

v = \frac{d x}{d t}

or,

v = A w c o s w t

result

Acceleration as a function of displacement

Acceleration

a = \frac{d v}{d t}

a = - A w^{2} s i n w t

a = - w^{2} x

Example:
Acceleration-displacement graph of a particle executing SHM is as shown in the figure. What is the time period of oscillation (in sec) ?

Solution:
In SHM

a = - ω^{2} x

a = \frac{- K}{m} x

so, from graph

- \frac{K}{m} = - 1

(∵ s l o p e i s - 1)

\frac{K}{m} = 1

Time period

= 2 π \frac{m}{K}

= 2 π \frac{1}{1}

= 2 π

example

Velocity as a function of displacement

General equation of SHM for displacement in a simple harmonic motion is:

x = A s i n w t

By definition,

v = \frac{d x}{d t}

or,

v = A w c o s w t

... (1)

Since

s i n^{2} w t + c o s^{2} w t = 1

From equation (1).

v = w A^{2} - x^{2}

result

Find acceleration from displacement as a function of time

In SHM for displacement has a time dependence equation in the form

x = A s i n w t

By definition,

v = \frac{d x}{d t}

or,

v = A w c o s w t

Acceleration is given by

a = \frac{d v}{d t}

a = - A w^{2} s i n w t

example

Interpret direction and magnitude of velocity for different positions in SHM

Example: Two particles

P

and

Q

describe simple harmonic motions of same period, same amplitude, along the same line about the same equilibrium position

O

. When

P

and

Q

are on opposite sides of

O

at the same distance from

O

they have the same speed of

1.2 m / s

in the same direction, when their displacements are the same they have the same speed of

1.6 m / s

in opposite directions. Find the maximum velocity in

m / s

of either particle?

Solution:
From the image,

ϕ = - θ + ω t

.............................................

(1)

180 - ϕ = θ + ω t

......................................

(2)

substituting value of

ϕ

from equn

(1)

(2)

, we will get

ω t = 90

.
That means that particles will have to rotate by

9 0^{0}

in order to be again at same displacement.
Hence from

(1),

we get

ϕ = - θ + 90

now from figure,

y = A sin θ

.....................................................

(3)

and

y^{'} = A sin ϕ = A sin (- θ + 90) = A cos θ = A 1 - sin^{2} θ

From eqn

(3)

y^{'} = A^{2} - y^{2}

....................................................

(4)

Initially magnitude of velocity,

= 1.2 m / s

Therefore

1.2 = ω A^{2} - y^{2}

......................

(5)

and Finally

1.6 = ω A^{2} - y^{' 2}

that gives,

1.6 = ω y

(

using eqn

(4))

......................

(6)

Putting

(6)

in eqn

(5)

after simplifying,we will get ,

ω A = 2 m / s

, is the magnitude of maximum velocity which occurs

at mean position.

result

Maximum and Minimum acceleration in SHM

General equation of acceleration is:

a = - A w^{2} s i n w t

From this expression one could infer directly that acceleration is maximum for:

w t = 90

( at the extreme position)
Acceleration is minimum for:

w t = 0

( at the mean position)

shortcut

Find maximum and minimum speed in SHM from velocity as a function of time

General equation of SHM for displacement in a simple harmonic motion is:

x = A s i n w t

By definition,

v = \frac{d x}{d t}

or,

v = A w c o s w t

... (1)

Clearly from equation (1) maximum velocity will be:

v = A w

for

w t = 0

and

v = 0

for

w t = 90

Which basically means velocity is maximum at the mean position and zero at the extreme position.

diagram

Draw a plot of acceleration as a function of time

Above graph clearly depicts variation of acceleration with time in a simple harmonic motion.

result

Compare plots of velocity and displacement as a function of time

In SHM in mean position magnitude of velocity is maximum and in extreme position velocity is zero. The above graphs of velocity and displacement depicts it clearly.

definition

Angular Acceleration as a function of time

α = - w^{2} θ

Angular acceleration as a function of time in SHM:

α = - w^{2} θ_{o} s i n w t

Since,

θ = θ_{o} s i n w t

result

Angular Velocity as a Function of Time in SHM

In angular SHM equation of motion is given by:

τ = - k θ

General equation for angular displacement:

θ = θ_{o} s i n (w t + ϕ)

Angular velocity

= \frac{d θ}{d t}

or, Angular velocity =

θ_{o} w c o s (w t + ϕ)

where

w = \frac{2 π}{T}

Where

T

is time period of SHM.

diagram

Draw a plot of acceleration as a function of distance

Since,

a = - w^{2} x

a

x

plot essentially represents a straight line with slope

- w^{2}

diagram

Direction of acceleration in SHM

Acceleration is given by

a = - ω^{2} x

Direction of acceleration is opposite to the direction of displacement.

formula

Velocity in SHM as a function of time

result

Acceleration as a function of displacement

example

Velocity as a function of displacement

result

Find acceleration from displacement as a function of time

example

Interpret direction and magnitude of velocity for different positions in SHM

result

Maximum and Minimum acceleration in SHM

shortcut

Find maximum and minimum speed in SHM from velocity as a function of time

diagram

Draw a plot of acceleration as a function of time

result

Compare plots of velocity and displacement as a function of time

definition

Angular Acceleration as a function of time

result

Angular Velocity as a Function of Time in SHM

diagram

Draw a plot of acceleration as a function of distance

diagram

Direction of acceleration in SHM

REVISE WITH CONCEPTS

Graphical Representation of Displacement, Velocity and Acceleration

Quick Summary With Stories

Velocity and Acceleration in SHM

A body oscillates with SHM according to the equation x=5.0cos(2πt+π). At time t=1.5s, its displacement, speed and acceleration respectively is:

A particle is executing SHM along a straight line. Its velocities at distances x1​ and x2​ from the mean position are V1​ and V2​, respectively. Its time period is:

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A simple pendulum performs simple harmonic motion about x=0 with an amplitude ′a′ and time period ′T′. The speed of the pendulum at x=2a​ will be:

A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be:

If the displacement (x) and velocity (v) of a particle executing S.H.M. are related through the expression 4v2=25−x2, then its time period is :

Two simple harmonic motions are represented by the equations y1​=0.1sin(100πt+3π​) and y2​=0.1cosπt. The phase difference of the velocity of particle 1 with respect to velocity of particle 2 at t=0 is

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance 32A​ from equilibrium position. The new amplitude of the motion is.

The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is

The x−t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t=34​ s is

A body oscillates with SHM according to the equation $x = 5.0 c o s (2 π t + π)$ . At time $t = 1.5 s$ , its displacement, speed and acceleration respectively is:

A particle is executing SHM along a straight line. Its velocities at distances $x_{1}$ and $x_{2}$ from the mean position are $V_{1}$ and $V_{2}$ , respectively. Its time period is:

A particle executes linear simple harmonic motion with an amplitude of $3 c m$ . When the particle is at $2 c m$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $^{'} a^{'}$ and time period $^{'} T^{'}$ . The speed of the pendulum at $x = \frac{a}{2}$ will be:

A particle is executing a simple harmonic motion. Its maximum acceleration is $α$ and maximum velocity is $β$ . Then, its time period of vibration will be:

If the displacement (x) and velocity (v) of a particle executing S.H.M. are related through the expression $4 v^{2} = 25 - x^{2}$ , then its time period is :

Two simple harmonic motions are represented by the equations $y_{1} = 0.1 sin (100 π t + \frac{π}{3})$ and $y_{2} = 0.1 cos π t$ . The phase difference of the velocity of particle $1$ with respect to velocity of particle $2$ at $t = 0$ is

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance $\frac{2 A}{3}$ from equilibrium position. The new amplitude of the motion is.

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7 m m$ , is $4.4 m / s$ . The period of oscillation is

The $x - t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t = \frac{4}{3} s$ is