The kinetic energy of a simple harmonic oscil | vedclass

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

Question

(A) The angular frequency of the kinetic energy oscillation is given as $\omega_k = 176 \ rad/s$. For a simple harmonic oscillator,the kinetic energy

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) The angular frequency of the kinetic energy oscillation is given as $\omega_k = 176 \ rad/s$. For a simple harmonic oscillator,the kinetic energy $K = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi)$. Using the identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$,we get $K = \frac{1}{4} m \omega^2 A^2 (1 - \cos(2\omega t + 2\phi))$. This shows that the kinetic energy oscillates with an angular frequency $\omega_k = 2\omega$,where $\omega$ is the angular frequency of the oscillator. Given $\omega_k = 176 \ rad/s$,we have $2\omega = 176 \ rad/s$,which implies $\omega = 88 \ rad/s$. The frequency of the oscillator is $f = \frac{\omega}{2\pi} = \frac{88}{2 	imes (22/7)} = \frac{88 	imes 7}{44} = 2 	imes 7 = 14 \ Hz$.

The kinetic energy of a simple harmonic oscillator is oscillating with an angular frequency of $176 \ rad/s$. The frequency of this simple harmonic oscillator is . . . . . . $Hz$. $\left[\text{take } \pi=\frac{22}{7}\right]$

Similar Questions

The kinetic energy of a particle executing $S.H.M.$ is $16 \, J$ when it is at its mean position. If the mass of the particle is $0.32 \, kg$,then what is the maximum velocity of the particle in $m/s$?

$A$ particle executes a linear $S.H.M.$ In two of its positions,the velocities are $V_1$ and $V_2$,and the accelerations are $a_1$ and $a_2$ respectively $(0 < a_1 < a_2)$. The distance between the positions is

$A$ particle performs linear $S.H.M.$ At a particular instant,velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant,velocity is $v$ and acceleration is $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions is

An object undergoing simple harmonic motion takes $0.5 \text{ s}$ to travel from one point of zero velocity to the next such point. The angular frequency of the motion is,

$A$ particle performing linear $S.H.M.$ of amplitude $0.1 \,m$ has displacement $0.02 \,m$ and acceleration $0.5 \,m/s^2$. The maximum velocity of the particle in $m/s$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module