Question

Three identical springs of force constant $100\text{N/m}$ attached to mass as shown in figure.


Calculate the time period of oscillation after a small displacement of mass $m=1\text{kg}$ towards spring $3.$

• A. $0.5\text{s}$
• B. $0.4\text{s}$
• C. $0.7\text{s}$
• D. $1.0\text{s}$

Answer 1

The correct option is B $0.4\text{s}$

Suppose block displaced $x\text{m}$ along the direction of spring $3$.


$kx=F_3$

$k{x}^{\prime }=F_1$

$k{x}^{\prime \prime }=F_2$

Elongation in spring $1$ is $x,$ and elongation in spring $2$ is $x^{\prime \prime }$

$x^{\prime }=x\cos {45}^{\circ }=\frac{x}{\sqrt{2}}$

$x^{\prime \prime }=x\sin {45}^{\circ }=\frac{x}{\sqrt{2}}$

$\therefore \left|{\overrightarrow{F}}_1\right|=kx,$

$\left|{\overrightarrow{F}}_2\right|=k\frac{x}{\sqrt{2}},\left|{\overrightarrow{F}}_3\right|=k\frac{x}{\sqrt{2}}$

$\therefore \left|\overrightarrow{F}\right|=\left|{\overrightarrow{F}}_1\right|\cos 45+\left|{\overrightarrow{F}}_2\right|\sin 45+\left|{\overrightarrow{F}}_3\right|$

$\Rightarrow k_{eq}x=\frac{kx}{2}+\frac{kx}{2}+kx$

Therefore, the equivalent force constant is $k_{eq}=2k$.

Time period, $T=2\pi \sqrt{\frac{m}{k_{eq}}}$

$=2\pi \sqrt{\frac{1}{2k}}$

$\therefore T=2\pi \sqrt{\frac{1}{200}}=0.44\text{s}$

Hence, option $\left(\text{B}\right)$ is correct.