Physics - SHM Question with Solution | TestHub

The potential energy is given by $U=10-10\cos (6x)$.

To find the force, we differentiate the potential energy with respect to $x$:

$$F=-\frac{dU}{dx}=-\frac{d}{dx}(10-10\cos (6x))=-60\sin (6x)$$

For small oscillations, we can use the small angle approximation $\sin (\theta )\approx \theta$. Therefore,

$$F\approx -60(6x)=-360x$$

Since $F=ma$, we have $ma=-360x$, where $m=1$ kg. Thus, the acceleration is

$$a=-360x$$

Comparing this with the standard form of simple harmonic motion $a=-{\omega }^{2}x$, we find that

$${\omega }^2=360$$

$$\omega =\sqrt{360}=\sqrt{36\times 10}=6\sqrt{10}\approx 6\pi$$

The period of oscillation is given by

$$T=\frac{2\pi }{\omega }=\frac{2\pi }{6\pi }=\frac{1}{3}$$

Given that the period is $1/n$ seconds, we have $1/n=1/3$, so $n=3$.

FINAL ANSWER: 3