Question

A weightless spring of length 60 cm and force constant $100N{m}^{-1}$ is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. A mass of 0.25 kg is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is:

• A. $\frac{\pi }{20}s$
• B. $\frac{\pi }{10}s$
• C. $\frac{\pi }{5}s$
• D. $\frac{\pi }{\sqrt{200}}s$

Answer 1

The correct option is A $\frac{\pi }{20}s$

$k_{eq}=k_1+k_2$

Now, $k_1{l}_1=k_2{l}_2=kl$ where $k=100$ and $l=60$

$k_1=\frac{kl}{l_1}=\frac{60\times 100}{30}=200$

$k_2=\frac{kl}{l_2}=\frac{60\times 100}{30}=200$

So, $k_{eq}=k_1+k_2=200+200=400$

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

$=2\pi \sqrt{\frac{0.25}{400}}$

$=\frac{\pi }{20}s$