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]\\)

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}\\)

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\\).

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.

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 \\)

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.

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

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\\)

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

(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\\)

An object of mass 0.2 kg executes SHM along x-axis with a frequency of \\(\dfrac { 25}{ \pi }\\) Hz. At the position x=0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation will be

At \\(x=0.01\ m\\), the Total energy \\(=K.E+P.E\\)\\(T.E=0.5\ J+0.4\ J=0.9\ J\\)we know that \\(T.E\\) of \\(SHM=\dfrac {1}{2} mw^2 A^2\\)\\(m=0.2,\ w=2\pi f=50\\)\\(\Rightarrow \ \dfrac {1}{2}mw^2 A^2=0.9\\)\\(\dfrac {1}{2}\times 0.2\times (50)^2 A^2 =0.9\\)\\(A=0.06\ m\ (A)\\)

Introduction to Simple Harmonic Motion | Definition, Examples, Diagrams

Q: An object of mass 0.2 kg executes SHM along x-axis with a frequency of \\(\dfrac { 25}{ \pi }\\) Hz. At the position x=0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation will be

At \\(x=0.01\ m\\), the Total energy \\(=K.E+P.E\\)\\(T.E=0.5\ J+0.4\ J=0.9\ J\\)we know that \\(T.E\\) of \\(SHM=\dfrac {1}{2} mw^2 A^2\\)\\(m=0.2,\ w=2\pi f=50\\)\\(\Rightarrow \ \dfrac {1}{2}mw^2 A^2=0.9\\)\\(\dfrac {1}{2}\times 0.2\times (50)^2 A^2 =0.9\\)\\(A=0.06\ m\ (A)\\)

definition

Time-period from equation of SHM

Equation of SHM:

F = - m ω^{2} x

Angular frequency

= ω

Frequency

f = \frac{ω}{2 π}

Time-period

T = \frac{2 π}{ω}

Example:
Acceleration-displacement graph of a particle executing SHM is as shown in the figure. Find the time period of oscillation.
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

definition

Simple Harmonic Motion

definition

Applications of SHM

Some applications of SHM are:

Simple harmonic motion of a pendulum is used for the measurement of time.
Tuning of the musical instrument is done with the vibrating tuning fork which executes simple harmonic motion.
Wave is a consequence of simple harmonic motion. Study of waves is indirectly the study of simple harmonic motion.
Molecules are in simple harmonic motion. This study is called vibration spectroscopy.

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Characteristics of SHM

A restoring force must act on the body.
Body must have acceleration in a direction opposite to the displacement and the acceleration must be directly proportional to displacement.
The system must have inertia (mass).
SHM is a type of oscillatory motion.
It is a particular case of preodic motion.
It can be represented by a simple sine or cosine function

Introduction to Simple Harmonic Motion

definition

Time-period from equation of SHM

definition

Simple Harmonic Motion

definition

Applications of SHM

definition

Characteristics of SHM

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