Question

In a gravity free space, a particle is connected with four springs as shown below. If the particle is pulled down gently and left, then find the period of oscillation of the particle.
$(\text{Given}:\theta ={45}^{\circ },\beta ={30}^{\circ })$

• A. $T=2\pi \sqrt{\frac{m}{4k}}$
• B. $T=2\pi \sqrt{\frac{m}{6k}}$
• C. $T=2\pi \sqrt{\frac{3m}{4k}}$
• D. $T=2\pi \sqrt{\frac{m}{2k}}$

Answer 1

The correct option is D $T=2\pi \sqrt{\frac{m}{2k}}$

Suppose the particle is displaced by a distance $x$ downward. Then, the upper two springs will be elongated and lower springs will be compressed. The force will act as shown in the FBD given below:


Let $y$ be the elongation in the upper springs. The net restoring force by the upper springs is


$F^{\prime }=F_1\cos (90-\theta )+F_2\cos (90-\theta )$
By symmetry, we can say that, $F_1=F_2=F\left(\text{say}\right)$
$\Rightarrow F^{\prime }=2F\sin \theta .....\left(1\right)$

Let the lower springs be compressed by $y^{{}^{\prime }}$ each.
The net restoring force by the two lower springs is given by
$F^{\prime \prime }=F_3\cos (90-\beta )+F_4\cos (90-\beta )$
By symmetry, we can say that $F_3=F_4=f\left(\text{say}\right)$
$\Rightarrow F^{\prime \prime }=2f\sin \beta ......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get
$F_{\text{net}}=2F\sin \theta +2f\sin \beta$
$F_{\text{net}}=-(2ky\sin \theta +4k{y}^{\prime }\sin \beta )$
From the first diagram, we can deduce that
Elongation in upper spring $y=x\cos (90-\theta )=x\sin \theta$
Elongation in lower spring $y^{\prime }=x\cos (90-\beta )=x\sin \beta$
Therefore,
$F_{\text{net}}=-(2kx{\sin }^2\theta +4kx{\sin }^2\beta )$
From the data given in the question,
$F_{\text{net}}=-[2kx\frac{1}{2}+4kx\frac{1}{4}]$
$F_{\text{net}}=-2kx\Rightarrow a=\frac{-2kx}{m}$
Comparing this with $a=-{\omega }^{2}x$, we get
$\omega =\sqrt{\frac{2k}{m}}\Rightarrow T=2\pi \sqrt{\frac{m}{2k}}$
Thus, option (d) is the correct answer.