Physics - SHM Question with Solution | TestHub

The given equation is $4V^2=50-x^2$

Rearranging the terms, we get:

$V^2=\frac{50}{4}-\frac{x^2}{4}$

$V^2=\frac{1}{4}(50-x^2)$

Comparing this with the standard equation of velocity in S.H.M., $V^2={\omega }^2(A^2-x^2)$, we get:

${\omega }^2=\frac{1}{4}$

$\omega =\frac{1}{2}$ rad/s

The time period of oscillation is given by $T=\frac{2\pi }{\omega }$

$T=\frac{2\pi }{1/2}=4\pi$

Substituting $\pi =\frac{22}{7}$:

$T=4\times \frac{22}{7}=\frac{88}{7}$ s

Given that the time period is $\frac{x}{7}$ s, we find $x=88$.

Answer: 88