A particle is suspended from a vertical sprin | vedclass

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

Question

(A) The frequency of the $S.H.M.$ is $f = 5 \ Hz$.Angular frequency $\omega = 2\pi f = 2\pi 	imes 5 = 10\pi \ rad/s$.At the highest point of oscillat

Vedclass · Accepted Answer

$\frac{1}{\pi} \ m/s$

Vedclass · Answer

(A) The frequency of the $S.H.M.$ is $f = 5 \ Hz$.Angular frequency $\omega = 2\pi f = 2\pi 	imes 5 = 10\pi \ rad/s$.At the highest point of oscillation,the spring is unstretched,which means the extension at the equilibrium position is equal to the amplitude $A$ of the oscillation.Thus,$A = \Delta l = \frac{mg}{k}$.Since $\omega^2 = \frac{k}{m}$,we have $\frac{g}{\omega^2} = \frac{mg}{k} = A$.Substituting the values: $A = \frac{10}{(10\pi)^2} = \frac{10}{100\pi^2} = \frac{1}{10\pi^2} \ m$.The maximum speed is given by $V_{\max} = A\omega$.$V_{\max} = \left(\frac{1}{10\pi^2}ight) 	imes (10\pi) = \frac{1}{\pi} \ m/s$.

$A$ particle is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 \ Hz$. The spring is unstretched at the highest point of oscillation. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)

Similar Questions

The correct stability order for the following carbocations is:

Lungs do not collapse between breaths and some air always remains in the lungs which can never be expelled because

$\frac{(1+i)^{2011}}{(1-i)^{2009}}$ is equal to

$A$ point charge $+q$ is placed at the centre of a cube of side $L$. The electric flux emerging from the cube is

In a $\triangle ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then $\triangle ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module