Q: A particle executes simple harmonic motion with a frequency \$'f'\$. The frequency with which its kinetic energy oscillates is?

Let, \$x=A\sin\omega t\$\$v=\cfrac {dx}{dt}=A\omega \cos \omega t\$Kinetic energy, \$K=\cfrac {1}{2}mv^2\$\$\implies K=\cfrac {1}{2}m\omega ^2A^2\cos^2\omega t\$\$\implies K=\cfrac {1}{2}m\omega^2A^2\left (\cfrac {1+cos2\omega t}{2}\right)\$\$\implies K=\cfrac {1}{4}m\omega^2A^2(1+\cos^2\omega t)\$\$\therefore \omega_K=2\omega\$\$\therefore\$ Frequency of oscillation of K.E \$=2f\quad \left[f=\cfrac {\omega}{2\pi}\right]\$

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

\$\begin{array}{l} { \theta _{ i } }=\dfrac { \pi }{ 2 } \\ { \theta _{ f } }=\dfrac { { 5\pi } }{ 6 } \\ \dfrac { { 2\pi } }{ T } t=\dfrac { { 5\pi } }{ 6 } -\dfrac { \pi }{ 2 } \\ =\dfrac { { 5\pi -3\pi } }{ 6 } \\ \dfrac { { 2\pi } }{ T } t=\dfrac { { 2\pi } }{ 6 } \\ t=\dfrac { T }{ 6 } \end{array}\$

Q: A simple harmonic motion is represented by \$x = 10\sin (20t + 0.5)\$.Write down its amplitude, angular frequency, frequency, time period and initial phase if displacement is measured in metres and time in second.

\$x = 10\sin (20t + 0.5) .....(1)\$Equation of SHM\$x = A \sin (\omega t + \phi) ..... (2)\$\$A = amplitude\$\$\omega =\$ angular frequency\$\phi =\$ initial phaseComparing coefficients of \$(1)\$ and \$(2)\$\$A = 10\ m\$\$\omega = 20\ rad\ s^{-1}\$\$\phi = 0.5\ rad\$\$\omega = 2\pi f\$\$f = \dfrac {\omega}{2\pi} = \dfrac {20}{2\times 3.14} = 3.18\ Hz\$\$T = \dfrac {2\pi}{\omega} = 2\pi \dfrac {2\times 3.14}{20} = 0.314\ s\$.

Q: The equation of motion of a particle executing SHM is \$(\dfrac{d^2x}{dt^2})+kx=0\$. The time period of the particle will be ?

The equation of motion,\$\dfrac{d^2x}{dt^2}+kx=0\$\$a=-kx\$\$\omega ^2=k\$ (\$a=-\omega ^2 x\$)\$\omega =\sqrt{k}\$\$\dfrac{2\pi}{T}=\sqrt{k}\$Time period,\$T=\dfrac{2\pi}{\sqrt{k}}\$The correct option is A.

Q: A simple pendulum completes 40 oscillation in a minute .Find its (a) frequency (b) time period.

Given Number of Oscillations, \$N=40\$ Time \$t=1\,\min \$ Frequency \$f=\dfrac{N}{t}=\dfrac{40}{60}=0.667\,Hz\$ Time period,\$T=\dfrac{1}{f}=1.5\,\sec \$ Hence, frequency is \$0.667\,Hz\$ and time period is \$1.5\,\,\sec \$

Q: A simple harmonic motion is represented by \$y = 5 ( \sin 3 \pi t + \sqrt { 3 } \cos 3 \pi t ) \mathrm { cm }\$ The amplitude and time period of the motion are :

\$y=5(\sin3\pi t+\sqrt{3}\cos 3\pi t)\$\$y=5\times 2\left(\dfrac{1}{2}\sin 3\pi t+\dfrac{\sqrt{3}}{2}\cos 3\pi t\right)\$\$y=10\left(\cos \dfrac{\pi}{3}\sin 3\pi t-\sin \dfrac{\pi}{3}\cos 3\pi t\right)\$\$y=10\sin\left(3\pi t+\dfrac{\pi}{3}\right)\$Comparing it with \$y=A\sin(\omega t+\phi)\$\$\Rightarrow A=10\$ i.e., amplitude is \$10\$cm\$\Rightarrow T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{3\pi}=\dfrac{2}{3}\$ sec. i.e, time period is \$\dfrac{2}{3}\$ sec.

Q: A simple harmonic motion has an amplitude \$A\$ and time period \$T\$. The time required by it to travel \$x = A\$ to \$x = \dfrac {A}{2}\$ is:

\$x=A\cos\omega t\$ \$t=0,x=A\$At \$x=\dfrac{A}{2} \$,\$\dfrac{A}{2}=A\cos\omega t\$\$\omega t=\dfrac{\pi}{3}\$\$\dfrac{2\pi}{T}t=\dfrac{\pi}{3}\$\$t=\dfrac{T}{6}\$ where T is time period

Q: A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is :

\$V_{max}=a\omega \Rightarrow f=\dfrac{\omega }{2\pi }=\dfrac{V_{max}}{2\pi a}=\dfrac{31.4}{2\pi \times 5}=\dfrac{31.4}{10\pi }= 1Hz\$

Q: The displacement of a particle in S.H.M. is indicated by equation \$y\ = \ 10\ sin(20t\ +\ \pi/3)\$ where \$y\$ is in metres. The value of time period of vibration will be (in seconds):

\$y=10\sin(20t+\dfrac{\pi}{3})\$Comparing with below equation, \$y=A\sin(\omega t+\phi)\$ where \$\omega=\$angular velocity=20Time period \$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\omega}=\dfrac{\pi}{10}\$

Question 1

A particle executes simple harmonic motion with a frequency $'f'$. The frequency with which its kinetic energy oscillates is?

Accepted Answer

Let, $x=A\sin\omega t$$v=\cfrac {dx}{dt}=A\omega \cos \omega t$Kinetic energy, $K=\cfrac {1}{2}mv^2$$\implies K=\cfrac {1}{2}m\omega ^2A^2\cos^2\omega t$$\implies K=\cfrac {1}{2}m\omega^2A^2\left (\cfrac {1+cos2\omega t}{2}\right)$$\implies K=\cfrac {1}{4}m\omega^2A^2(1+\cos^2\omega t)$$\therefore \omega_K=2\omega$$\therefore$ Frequency of oscillation of K.E $=2f\quad \left[f=\cfrac {\omega}{2\pi}\right]$

Question 2

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

Accepted Answer

$\begin{array}{l} { \theta _{ i } }=\dfrac { \pi  }{ 2 }  \ { \theta _{ f } }=\dfrac { { 5\pi  } }{ 6 }  \ \dfrac { { 2\pi  } }{ T } t=\dfrac { { 5\pi  } }{ 6 } -\dfrac { \pi  }{ 2 }  \ =\dfrac { { 5\pi -3\pi  } }{ 6 }  \ \dfrac { { 2\pi  } }{ T } t=\dfrac { { 2\pi  } }{ 6 }  \ t=\dfrac { T }{ 6 }  \end{array}$

Question 3

A simple harmonic motion is represented by $x = 10 sin (20 t + 0.5)$ .
Write down its amplitude, angular frequency, frequency, time period and initial phase if displacement is measured in metres and time in second.

Accepted Answer

$x = 10\sin (20t + 0.5) .....(1)$Equation of SHM$x = A \sin (\omega t + \phi) ..... (2)$$A = amplitude$$\omega =$ angular frequency$\phi =$ initial phaseComparing coefficients of $(1)$ and $(2)$$A = 10\ m$$\omega = 20\ rad\ s^{-1}$$\phi = 0.5\ rad$$\omega = 2\pi f$$f = \dfrac {\omega}{2\pi} = \dfrac {20}{2\times 3.14} = 3.18\ Hz$$T = \dfrac {2\pi}{\omega} = 2\pi \dfrac {2\times 3.14}{20} = 0.314\ s$.

Question 4

The equation of motion of a particle executing SHM is $(\dfrac{d^2x}{dt^2})+kx=0$. The time period of the particle will be ?

Accepted Answer

The equation of motion,$\dfrac{d^2x}{dt^2}+kx=0$$a=-kx$$\omega ^2=k$    ($a=-\omega ^2 x$)$\omega =\sqrt{k}$$\dfrac{2\pi}{T}=\sqrt{k}$Time period,$T=\dfrac{2\pi}{\sqrt{k}}$The correct option is A.

Question 5

A simple pendulum completes 40 oscillation in a minute .Find its (a) frequency (b) time period.

Accepted Answer

Given&#10;&#10;Number of Oscillations, $N=40$&#10;&#10;Time $t=1\,\min $&#10;&#10;Frequency $f=\dfrac{N}{t}=\dfrac{40}{60}=0.667\,Hz$&#10;&#10;Time period,$T=\dfrac{1}{f}=1.5\,\sec $&#10;&#10;Hence, frequency is $0.667\,Hz$ and time period is $1.5\,\,\sec&#10;$

Question 6

A simple harmonic motion is represented by $y = 5 ( \sin 3 \pi t + \sqrt { 3 } \cos 3 \pi t ) \mathrm { cm }$  The amplitude and time period of the motion are :

Accepted Answer

$y=5(\sin3\pi t+\sqrt{3}\cos 3\pi t)$$y=5\times 2\left(\dfrac{1}{2}\sin 3\pi t+\dfrac{\sqrt{3}}{2}\cos 3\pi t\right)$$y=10\left(\cos \dfrac{\pi}{3}\sin 3\pi t-\sin \dfrac{\pi}{3}\cos 3\pi t\right)$$y=10\sin\left(3\pi t+\dfrac{\pi}{3}\right)$Comparing it with $y=A\sin(\omega t+\phi)$$\Rightarrow A=10$ i.e., amplitude is $10$cm$\Rightarrow T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{3\pi}=\dfrac{2}{3}$ sec. i.e, time period is $\dfrac{2}{3}$ sec.

Question 7

A simple harmonic motion has an amplitude $A$ and time period $T$. The time required by it to travel $x = A$ to $x = \dfrac {A}{2}$ is:

Accepted Answer

$x=A\cos\omega t$  $t=0,x=A$At $x=\dfrac{A}{2} $,$\dfrac{A}{2}=A\cos\omega t$$\omega t=\dfrac{\pi}{3}$$\dfrac{2\pi}{T}t=\dfrac{\pi}{3}$$t=\dfrac{T}{6}$ where T is time period

Question 8

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is :

Accepted Answer

$V_{max}=a\omega \Rightarrow f=\dfrac{\omega }{2\pi }=\dfrac{V_{max}}{2\pi a}=\dfrac{31.4}{2\pi \times 5}=\dfrac{31.4}{10\pi }= 1Hz$

Question 9

A particle executes simple harmonic motion between x = - A  and x = + A .The time taken by it to go from O to A /2 is T$_{1}$ and to go from  A /2 to  A is T$_{2}$. Then

Accepted Answer

(a) The equation of SHM, when it starts swinging from the mean position, is $y=Asin\omega{t}$ .At mean position, y=0,   t=0      .....(1)$\dfrac{A}{2}=Asin\omega{t_2}$    or  $sin\omega{t_2}=\dfrac{1}{2}=sin\dfrac{\pi}{6}$      $\omega{t_2}=\dfrac{\pi}{6}$  or  $t_2=\dfrac{\pi}{6\omega}$    .....(2)At y=A time for travel from 0 to A=$t_3$A=Asin$\omega{t_3}$or sin$\omega{t_3}=1=sin\dfrac{\pi}{2}$  or $t_3=\dfrac{\pi}{2\omega}$      .....(3)$T_1= t_2-t_1=\dfrac{\pi}{6\omega}$     .....(4)$T_2=t_3-t_2=\dfrac{\pi}{2\omega}-\dfrac{\pi}{6\omega}=\dfrac{\pi}{3\omega}$    .....(5)$\dfrac{T_1}{T_2}={\dfrac{\dfrac{\pi}{6\omega}}{\dfrac{\pi}{3\omega}}}=\dfrac{3\omega}{6\omega}=\dfrac{1}{2}$ or $\dfrac{T_1}{T_2}=\dfrac{1}{2}$$T_1<T_2$

Question 10

The displacement of a particle in S.H.M. is indicated by equation $y\ = \ 10\ sin(20t\ +\ \pi/3)$ where $y$ is in metres. The value of time period of vibration will be (in seconds):

Accepted Answer

$y=10\sin(20t+\dfrac{\pi}{3})$Comparing with below equation, $y=A\sin(\omega t+\phi)$ where $\omega=$angular velocity=20Time period $T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\omega}=\dfrac{\pi}{10}$

Periodic motion and simple harmonic motion

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