For a simple harmonic motion with amplitude A | vedclass

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

Question

(A) The velocity $v$ of a particle in simple harmonic motion at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$.Given the an

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The velocity $v$ of a particle in simple harmonic motion 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 $x = \frac{A}{2}$,we substitute these values into the 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}}$$v = \frac{2\pi}{T} \cdot \frac{A\sqrt{3}}{2}$$v = \frac{\pi A \sqrt{3}}{T}$.

For a simple harmonic motion with amplitude $A$ and time period $T$,what is the velocity at $x = \frac{A}{2}$?

Similar Questions

The graph in the figure represents:

$A$ spring executes $S.H.M.$ with mass $10 \ kg$ attached to it. The force constant of the spring is $10 \ N/m$. If at any instant its velocity is $40 \ cm/s$,the displacement at that instant is (Amplitude of $S.H.M.$ $= 0.5 \ m$) (in $m$)

From the given displacement-time graph of an oscillating particle,find the maximum velocity of the particle.

Obtain the velocity of a particle executing simple harmonic motion $(SHM)$ by considering the projection of a particle undergoing uniform circular motion.

$A$ particle executes linear $S.H.M.$ with amplitude $A = 4 \text{ cm}$. The magnitude of velocity and acceleration is equal when it is at a distance $x = 3 \text{ cm}$ from the mean position. The time period of the oscillation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module