Question

The wedge of mass $m$ and inclination $\tan^{-1} 0.75$ with the horizontal is on a smooth horizontal surface while attached to a long ideal spring of force constant $k$. The spring is along a normal to the inclined face of the wedge. When the wedge is given a small displacement, the angular frequency $\omega$ of small oscillations satisfies $nm\omega^2 = k$. Find the integer closest to $18n$.

Answer 1

50

Answer 2

Explanation:

1. Let the plane’s inclination be $\theta$ where $\tan\theta=0.75$. Then, $\sin\theta=0.6$ and $\cos\theta=0.8$.
2. The spring is attached along the normal to the plane. Its horizontal projection is the absolute value of the x‐component of the unit normal. The unit normal is $$\mathbf{n}=(-\sin\theta,\cos\theta)$$ so its horizontal component is $|-\sin\theta|=\sin\theta=0.6$.
3. For a small horizontal displacement $\delta x$, the compression (or extension) of the spring is $\delta l=\delta x\sin\theta$. Therefore, the spring force acting horizontally is $$F_x=k\,\delta l\,( \sin\theta)=k\,\delta x\sin^2\theta.$$
4. The effective horizontal spring constant is $k_{\rm eff}=k\sin^2\theta$. Thus, the angular frequency of oscillation is $$\omega^2=\dfrac{k_{\rm eff}}{m}=\dfrac{k\sin^2\theta}{m}.$$
5. Given that the oscillation frequency satisfies $$nm\,\omega^2=k,$$ substitute $\omega^2$ to obtain: $$nm\left(\frac{k\sin^2\theta}{m}\right)=n k\sin^2\theta=k.$$
6. Cancelling $k$ leads to: $$n\sin^2\theta=1\quad\Longrightarrow\quad n=\frac{1}{\sin^2\theta}=\frac{1}{0.36}\approx2.7777.$$
7. Thus, $18n\approx18\times2.7777\approx50$.

Answer: 50