JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 30)
The time period of simple harmonic motion of mass $$M$$ in the given figure is $$\pi \sqrt{\frac{\alpha M}{5 k}}$$, where the value of $$\alpha$$ is _________.
_31st_January_Evening_Shift_en_30_1.png)
Answer
12
Explanation
$$\mathrm{k}_{\mathrm{eq}}=\frac{2 \mathrm{k} \cdot \mathrm{k}}{3 \mathrm{k}}+\mathrm{k}=\frac{5 \mathrm{k}}{3}$$
Angular frequency of oscillation $$(\omega)=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{m}}}$$
$$(\omega)=\sqrt{\frac{5 \mathrm{k}}{3 \mathrm{~m}}}$$
Period of oscillation $$(\tau)=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{3 \mathrm{~m}}{5 \mathrm{k}}}$$
$$=\pi \sqrt{\frac{12 \mathrm{~m}}{5 \mathrm{k}}}$$
Comments (0)
