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: 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: 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 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: 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 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 \$

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

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 5

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 6

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 7

Define linear S.H.M. Obtain differential equation of linear S.H.M.

Accepted Answer

Linear SHM is the simplest Kind of oscillatory motion in which a body when displaced from its mean position, oscillates 'to and fro' about mean position and the restoring force (or acceleration) is always directed towards its mean position and its magnitude is directly proportional to the displacement from the mean position.Consider particle $P$ moving along the circumference of a circle of radius 'a' with a uniform angular speed of $\omega$ in the anticlockwise direction.Refer image .1Particle $P$ along the circumference of the circle has its projection particle on diameter $AB$ at point $M$. This projection particle follows linear SHM along line $AB$ when particle P rotates around the circle.Let the rotation start with initial angle $'\alpha'$ as shown above $(t=0)$In time 't' the angle between OP and X-axis will become $(\omega t+\alpha)$ as shown below:Refer image 2Now from $\Delta\,\,OPM=\dfrac{OM}{OP}\cos (\omega t+\alpha)$                         $\Rightarrow \dfrac{x}{a}=\cos(\omega t+\alpha)$                        $\Rightarrow x=a\cos (\omega t+\alpha)$               .........(1)Differentiating above equation with respect to time we get velocity$\Rightarrow v=\dfrac{dx}{dt}=-a\omega \sin (\omega t+\alpha)$                   .........(2)Differentiating again we get acceleration$\Rightarrow a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}-a\omega^2\cos(\omega t+\alpha)$$=-a\cos(\omega t+\alpha)\omega^2$$=-\omega^2x$                                  ...........(3)as $x=a\cos (\omega t+\alpha)$Equation (1),(2) and (3) are differential equations for linear SHM.

Question 8

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 9

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 10

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$

Introduction to Simple Harmonic Motion

Language

Rate

REVISE WITH CONCEPTS