The figure shows the variation of force actin | vedclass

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

Question

(B) For a particle executing simple harmonic motion,the force is given by $F = -kx$,where $k$ is the force constant.From the graph,at $x = 20\, cm = 0

Vedclass · Accepted Answer

$\left( \frac{5}{2\pi} \right)\, s^{-1}$

Vedclass · Answer

(B) For a particle executing simple harmonic motion,the force is given by $F = -kx$,where $k$ is the force constant.From the graph,at $x = 20\, cm = 0.2\, m$,the force $F = -2.0\, N$.Thus,the magnitude of the force constant is $k = \frac{|F|}{|x|} = \frac{2.0\, N}{0.2\, m} = 10\, N/m$.The mass of the particle is $m = 400\, g = 0.4\, kg$.The frequency of oscillation $f$ is given by the formula $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.Substituting the values,we get $f = \frac{1}{2\pi} \sqrt{\frac{10}{0.4}} = \frac{1}{2\pi} \sqrt{25} = \frac{5}{2\pi}\, s^{-1}$.

The figure shows the variation of force acting on a particle of mass $400\, g$ executing simple harmonic motion. The frequency of oscillation of the particle is

Similar Questions

For a particle executing simple harmonic motion, the displacement-time $(x-t)$ graph is as shown in the figure. The acceleration of the particle at $t=\frac{4}{3} \,s$ is

Two simple harmonic motions of angular frequency $100 \, rad \, s^{-1}$ and $1000 \, rad \, s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration is

Obtain instantaneous acceleration of a $SHM$ particle with the help of reference circle.

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

Write the formula of instantaneous acceleration of $SHM$ particle along $X$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module