Q: 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.

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: 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}\$

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: 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}\$

Q: 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 } \$

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}\$

Question 1

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:

Accepted Answer

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.

Question 2

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:

Accepted Answer

&#10;&#10;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;Using the&#10;expressions for velocity for SHM,$v_1 = \omega \sqrt{A^2&#10;-x_1^2}$$v_2 = \omega \sqrt{A^2 -x_2^2}$Squaring&#10;both,$v_1 ^2 = \omega ^2 ( A^2 -x_1 ^2)$ ....(3)$v_2 ^2 =&#10;\omega ^2 ( A^2 -x_2 ^2)$ ....(4)Subtracting (4) from (3), $v_1&#10;^2 - v_2 ^2 = \omega ^2 ( x_2 ^2 - x_1 ^2) $$ \Rightarrow &#10;4 \pi ^2 ( x_2 ^2 - x_1 ^2) = T^2 (v_1 ^2 - v_2 ^2 )$$&#10;\Rightarrow T = 2\pi \sqrt{\dfrac{ x_2 ^2 - x_1 ^2}{v_1 ^2 - v_2&#10;^2}}$ where we have used $\omega = \dfrac{2\pi}{T}$&#10;&#10;

Question 3

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

Accepted Answer

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 }  } $

Question 4

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:

Accepted Answer

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}$

Question 5

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:

Accepted Answer

For SHM, maximum acceleration, $a_{max}=\omega^2 a=\alpha$  &#10; where a is the amplitude of SHM.&#10;and&#10;maximum velocity, $v_{max}=\omega a=\beta$&#10;so, $\dfrac{\alpha}{\beta}=\dfrac{\omega^2&#10;a}{\omega a}=\omega$&#10;or $\dfrac{\alpha}{\beta}=\dfrac{2\pi}{&#10;T}$&#10;or $T=\dfrac{2\pi&#10;\beta}{\alpha}$

Question 6

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 :

Accepted Answer

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$

Question 7

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

Accepted Answer

$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$t = 0$              $= \dfrac{\pi}{3} - \dfrac {\pi}{2} = -\dfrac{\pi}{6}$

Question 8

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.

Accepted Answer

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 } $

Question 9

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

Accepted Answer

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 $T=9.99\times 10^{-3}=10\times 10^{-3}=0.01 s$

Question 10

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

Accepted Answer

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 $\mathrm{t}=4/3\ \mathrm{s}\mathrm{e}\mathrm{c}$,$\displaystyle \frac{\mathrm{d}^{2}\mathrm{x}}{\mathrm{d}\mathrm{t}^{2}}=-\dfrac{\sqrt{3}}{32}\pi^{2}\ \mathrm{c}\mathrm{m}/\mathrm{s}^{2}$

Revise with Concepts

Velocity and Acceleration in SHM

Graphical Representation of Displacement, Velocity and Acceleration

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Velocity and Acceleration in SHM

Velocity and Acceleration in SHM - Problems L1

Graphical representation of Displacement, Velocity and Acceleration

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

Revise with Concepts

Velocity and Acceleration in SHM

Graphical Representation of Displacement, Velocity and Acceleration

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Velocity and Acceleration in SHM

Velocity and Acceleration in SHM - Problems L1

Graphical representation of Displacement, Velocity and Acceleration

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