Question

On suspending a mass M from a spring of force constant K,frequency of vibration f is obtained If a second spring as shown in the figure. is arranged then the frequency will be :

• A. $f\sqrt{2}$
• B. $f/\sqrt{2}$
• C. 2f
• D. f

Answer 1

The correct option is B $f/\sqrt{2}$

Given,

Time period of SHM is one spring and mass is connected

Spring constant, $k$

Attached mass, $m$

So, frequency is, $f=2\pi \sqrt{\frac{k}{m}}$

As in the figure, if another spring is added in parallel.

Both spring identical, new spring constant for two parallel spring.

$k_{new}=k+k=2k$

Time period of oscillation,

$f_{new}=2\pi \sqrt{\frac{m}{k_{new}}}=2\pi \sqrt{\frac{m}{2k}}$

$f_{new}=\frac{1}{\sqrt{2}}2\pi \sqrt{\frac{m}{k}}=\frac{f}{\sqrt{2}}$

Hence, new frequency is $\frac{f}{\sqrt{2}}$