Question

When a mass $m$ is connected individually to two springs $S_1$ and $S_2$, the oscillation frequencies are $v_1$ and $v_2$. If the same mass is attached to the two springs as shown in Fig. the oscillation frequency would be:

• A. $\left(v_1+v_2\right)$
• B. $\sqrt{v_1^2+v_2^2}$
• C. ${\left(\frac{1}{v_1}+\frac{1}{v_2}\right)}^{-1}$
• D. $\sqrt{v_1^2-v_2^2}$

Answer 1

The correct option is B $\sqrt{v_1^2+v_2^2}$

When the mass is connected to the springs individually,

$v_1=\frac{1}{2\pi }\sqrt{\frac{k_1}{m}}-------\left(1\right)$

$v_2=\frac{1}{2\pi }\sqrt{\frac{k_2}{m}}--------\left(2\right)$

where $k_1$ and $k_2$ are the two spring constant Now, the block is connected with two springs.

Here equivalent spring constant $k_{eq}=k_1+k_2$

Time period of oscillation of the spring block system is

$T=2\pi \sqrt{\frac{m}{k_{eq}}}=2\pi \sqrt{\frac{m}{k_1+k_2}}$

$v=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{k_1+k_2}{m}}$

$v=\frac{1}{2\pi }[\frac{k_1}{m}+\frac{k_2}{m}]--------\left(3\right)$

(1)=>$\frac{k_1}{m}=4{\pi }^2{v}_1^2$

(2)=> $\frac{k_2}{m}=4{\pi }^2{v}_2^2$

=> $v=\frac{1}{2\pi }{[\frac{4{\pi }^2{v}_1^2}{1}+\frac{4{\pi }^2{v}_2^2}{1}]}^{\frac{1}{2}}$

$v=\sqrt{v_1^2+v_2^2}$