Under the action of a force F = -75 y, where | vedclass

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

Question

(D) Given the force equation $F = -75 y$. Comparing this with the standard $SHM$ force equation $F = -ky$, we get the force constant $k = 75 \,N/m$.Gi

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(D) Given the force equation $F = -75 y$. Comparing this with the standard $SHM$ force equation $F = -ky$, we get the force constant $k = 75 \,N/m$.Given mass $m = 3 \,kg$, the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{75}{3}} = \sqrt{25} = 5 \,rad/s$.The velocity at the mean position is the maximum velocity $V_{\max} = 2.5 \,m/s$.Since $V_{\max} = A\omega$, the amplitude $A = \frac{V_{\max}}{\omega} = \frac{2.5}{5} = 0.5 \,m$.The maximum acceleration is given by $a_{\max} = \omega^2 A$.Substituting the values, $a_{\max} = (5)^2 	imes 0.5 = 25 	imes 0.5 = 12.5 \,m/s^2$.

Under the action of a force $F = -75 y$, where $F$ is in Newton and $y$ is in meters, an object of mass $3 \,kg$ executes simple harmonic motion. If the velocity of the object at the mean position is $2.5 \,ms^{-1}$, the maximum acceleration of the object is (in $\,ms^{-2}$)

Similar Questions

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

The variation of displacement with time of a simple harmonic motion $(SHM)$ for a particle of mass $m$ is represented by $y = 2 \sin \left(\frac{\pi t}{2} + \phi\right) \text{ cm}$. The maximum acceleration of the particle 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:

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

The maximum acceleration of a particle in $SHM$ is made two times keeping the maximum speed constant. This is possible when:

Vedclass Test Series

Exam Paper Generator

Online Exam Module