A body is executing simple harmonic motion wi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The displacement equation for a simple harmonic motion $(SHM)$ with frequency $n$ is given by:$x = A \sin(\omega t) = A \sin(2 \pi n t)$The potent

Vedclass · Accepted Answer

$2n$

Vedclass · Answer

(C) The displacement equation for a simple harmonic motion $(SHM)$ with frequency $n$ is given by:$x = A \sin(\omega t) = A \sin(2 \pi n t)$The potential energy $(U)$ of the body is given by:$U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(2 \pi n t)$Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos(2 	heta)}{2}$,we get:$U = \frac{1}{2} k A^2 \left[ \frac{1 - \cos(2 \cdot 2 \pi n t)}{2} ight] = \frac{1}{4} k A^2 [1 - \cos(4 \pi n t)]$The frequency of the potential energy is the coefficient of $2 \pi t$ in the cosine term,which is $2n$.

$A$ body is executing simple harmonic motion with frequency $n$,the frequency of its potential energy is:

Similar Questions

The ratio of kinetic energy to the potential energy of a particle executing $SHM$ at a distance equal to half its amplitude,the distance being measured from its equilibrium position,is:

An object executes $SHM$ along the $x$-axis with an amplitude of $0.06 \,m$. At a certain distance '$x$' metres from the mean position,it has a kinetic energy of $10 \,J$ and a potential energy of $8 \,J$. The distance '$x$' will be: (in $\,m$)

Write the maximum velocity of a $SHM$ oscillator in terms of mechanical energy $E$ and mass of oscillator $m$.

$A$ particle performs $S.H.M.$ from the mean position. Its amplitude is $A$ and total energy is $E$. At a particular instant,its kinetic energy is $\frac{3E}{4}$. The displacement of the particle at that instant is:

On what factors does the mechanical energy of a simple harmonic oscillator depend,and on what factors does it not depend?

Vedclass Test Series

Exam Paper Generator

Online Exam Module