A particle is executing a simple harmonic mot | vedclass

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

Question

(A) For a particle in simple harmonic motion,the maximum acceleration is given by $a_{\max} = \omega^2 A = \alpha$ $...(1)$The maximum velocity is giv

Vedclass · Accepted Answer

$\frac{\alpha}{2\pi\beta}$

Vedclass · Answer

(A) For a particle in simple harmonic motion,the maximum acceleration is given by $a_{\max} = \omega^2 A = \alpha$ $...(1)$The maximum velocity is given by $v_{\max} = \omega A = \beta$ $...(2)$Dividing equation $(1)$ by equation $(2)$,we get:$\frac{\omega^2 A}{\omega A} = \frac{\alpha}{\beta}$$\omega = \frac{\alpha}{\beta}$Since angular frequency $\omega = 2\pi f$,where $f$ is the frequency of vibration:$2\pi f = \frac{\alpha}{\beta}$$f = \frac{\alpha}{2\pi\beta}$

$A$ particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then its frequency of vibration will be

Similar Questions

$A$ particle of mass $1 \ kg$ is moving in $SHM$ with a path length of $0.01 \ m$ and frequency of $50 \ Hz$. The maximum force in newton,acting on the particle is (in $\pi^2$)

In a simple harmonic motion,let $f$ be the acceleration and $T$ be the time period. If $x$ denotes the displacement,then the $|fT|$ vs. $x$ graph will look like,

The variation of the acceleration $a$ of a particle executing $S.H.M.$ with displacement $x$ is as shown in the figure. Which of the following graphs correctly represents this relationship?

In $S.H.M.$,maximum acceleration is at

$A$ particle of mass $10 \,g$ is undergoing $S.H.M.$ with an amplitude of $10 \,cm$ and a period of $0.1 \,s$. The maximum value of the force on the particle is about ............ $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module