Jee Main 2024 31 01 2024 Shift 2 - Question19
Question 19
The period of oscillation of system shown below is $\pi \sqrt{\frac{m}{5 k}}$ then $\alpha$ is
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Answer: (12)
Solution:
$k _{e q}=\frac{2 k \cdot k}{3 k}+k=\frac{5 k}{3}$
Angular frequency of oscillation $(\omega)=\sqrt{\frac{k _{e q}}{m}}$
$\omega=\sqrt{\frac{5 k}{3 m}}$
Period of oscillation $(\tau)=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{3 m}{5 k}}$
$ =\pi \sqrt{\frac{12 m}{5 k}} $
BETA
