Q 11-13-151JEE MainJEE Main 2017 (8 Apr)Easy
A $1$ kg block attached to a spring vibrates with a frequency of $1$ Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8$ kg block placed on the same table. So, the frequency of vibration of the $8$ kg block is:
Answer: (D) $\dfrac12$ Hz
$f = \dfrac{1}{2\pi}\sqrt{\dfrac km}$. Two springs in parallel give $2k$; the mass is $8$ times larger:
$$\frac{f'}{f} = \sqrt{\frac{2k/8}{k/1}} = \frac12 \;\Rightarrow\; f' = \frac12\ \text{Hz}$$
Solution by Sreeraj P, M.Sc Physics