Write the maximum velocity of a SHM oscillato | vedclass

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

Question

(B) The total mechanical energy $E$ of a $SHM$ oscillator is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude

Vedclass · Accepted Answer

$v_{\max} = \sqrt{\frac{2E}{m}}$

Vedclass · Answer

(B) The total mechanical energy $E$ of a $SHM$ oscillator is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.Since $k = m \omega^2$,we can write $E = \frac{1}{2} m \omega^2 A^2$.We know that the maximum velocity $v_{\max}$ of a $SHM$ oscillator is given by $v_{\max} = \omega A$.Substituting this into the energy equation: $E = \frac{1}{2} m (\omega A)^2 = \frac{1}{2} m v_{\max}^2$.Rearranging for $v_{\max}$,we get $v_{\max}^2 = \frac{2E}{m}$.Therefore,$v_{\max} = \sqrt{\frac{2E}{m}}$.

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

Similar Questions

$A$ particle executing simple harmonic motion along a straight line with an amplitude $A$,attains maximum potential energy when its displacement from the mean position equals

$A$ particle is performing $S.H.M.$ with energy of vibration $90 \,J$ and amplitude $6 \,cm$. When the particle reaches a distance of $4 \,cm$ from the mean position,it is stopped for a moment and then released. The new energy of vibration will be ........... $J$.

$A$ particle is vibrating in a simple harmonic motion with an amplitude of $4\, cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

The displacement of a simple harmonic motion of amplitude $6 \text{ cm}$ when its kinetic energy is equal to its potential energy is:

If $T$ is the period of $SHM$,then write the period of kinetic and potential energy.

Vedclass Test Series

Exam Paper Generator

Online Exam Module