A simple pendulum performs S.H.M about x = 0 | vedclass

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

Question

(B) The velocity $v$ of a particle performing $S.H.M$ at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$.Given the angular f

Vedclass · Accepted Answer

$\frac{\sqrt{3} \pi A}{T}$

Vedclass · Answer

(B) The velocity $v$ of a particle performing $S.H.M$ at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$.Given the angular frequency $\omega = \frac{2\pi}{T}$ and the displacement $x = \frac{A}{2}$.Substituting these values into the velocity formula:$v = \frac{2\pi}{T} \sqrt{A^2 - (\frac{A}{2})^2}$$v = \frac{2\pi}{T} \sqrt{A^2 - \frac{A^2}{4}}$$v = \frac{2\pi}{T} \sqrt{\frac{3A^2}{4}}$$v = \frac{2\pi}{T} \cdot \frac{\sqrt{3}A}{2}$$v = \frac{\sqrt{3} \pi A}{T}$.

$A$ simple pendulum performs $S.H.M$ about $x = 0$ with an amplitude $A$ and time-period $T$. The speed of the pendulum at $x = \frac{A}{2}$ will be

Similar Questions

For a simple harmonic motion with amplitude $A$ and time period $T$,at what position is the velocity half of the maximum velocity?

$A$ particle is performing simple harmonic motion with amplitude $A$ and angular velocity $\omega$. The ratio of maximum velocity to maximum acceleration is

$A$ particle executes $SHM$. Its velocities are $v_1$ and $v_2$ at displacements $x_1$ and $x_2$ from the mean position,respectively. The frequency of oscillation will be

If a simple pendulum oscillates with an amplitude of $50 \,mm$ and time period of $2 \,s$, then its maximum velocity is (in $\,ms^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module