Question

A particle of mass 0.1 kg is held between two rigid supports by two springs of force constant 8 N $m^{-1}$ and 2 N $m^{-1}$. If the particle is displaced along the direction of length of the springs, its frequency of vibration is

• A. $\frac{5}{\pi }Hz$
• B. $\frac{8}{\pi }Hz$
• C. $\frac{2}{\pi }Hz$
• D. $\frac{1}{\pi }Hz$

Answer 1

The correct option is A $\frac{5}{\pi }Hz$

Given that,

$m=0.1kg$

$k_1=8N/m$

$k_2=2N/m$

From the above figure,

$k_{eq}=k_1+k_2=10N/m$

$\therefore$ Frequency of the vibration $f=\frac{1}{2\pi }\sqrt{\frac{k_{eq}}{m}}$

$⟹f=\frac{1}{2\pi }\sqrt{\frac{10}{0.1}}=\frac{10}{2\pi }Hz$

$⟹f=\frac{5}{\pi }Hz$

So, correct option is $A$