Two simple harmonic motions of angular freque | vedclass

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

Question

(B) The maximum acceleration $a_{max}$ of a particle executing simple harmonic motion is given by the formula $a_{max} = \omega^2 A$,where $\omega$ is

Vedclass · Accepted Answer

$1: 10^2$

Vedclass · Answer

(B) The maximum acceleration $a_{max}$ of a particle executing simple harmonic motion is given by the formula $a_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the amplitude.Given that the amplitudes $A_1$ and $A_2$ are equal $(A_1 = A_2 = A)$,the ratio of the maximum accelerations is:$\frac{a_{max,1}}{a_{max,2}} = \frac{\omega_1^2 A}{\omega_2^2 A} = \left( \frac{\omega_1}{\omega_2} \right)^2$Substituting the given values $\omega_1 = 300 \ rad/s$ and $\omega_2 = 3000 \ rad/s$:$\frac{a_{max,1}}{a_{max,2}} = \left( \frac{300}{3000} \right)^2 = \left( \frac{1}{10} \right)^2 = \frac{1}{100} = 1: 10^2$Therefore,the correct option is $B$.

Two simple harmonic motions of angular frequency $300 \ rad/s$ and $3000 \ rad/s$ have the same amplitude. The ratio of their maximum accelerations is

Similar Questions

$Assertion :$ In $SHM$,acceleration is always directed towards the mean position.
$Reason :$ In $SHM$,the body has to stop momentarily at the extreme position and move back to the mean position.

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

The displacement-time graph of a particle executing $S.H.M.$ is as shown in the figure. The corresponding force-time graph of the particle is:

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

Obtain the force law for $SHM$ from the displacement of $SHM$ particle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module