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 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}\$

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 } \$

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>>Class 11
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>>Velocity and Acceleration in SHM
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Graphical Representation of Displacement, Velocity and Acceleration

Q: 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}\$

Q: 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 } } \$

Q: 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}\$

Q: 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\$

Q: 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\$

Q: 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}\$

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diagram

Plot of displacement as a function of time and interpret amplitude, phase and time period

From the above graph we can infer that:
Peak of the curve goes to height A, hence it's amplitude is A.
Time period is T, as it takes time T to complete one cycle.
Phase for this SHM is zero.

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Velocity and Acceleration in SHM

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