Oscillations - Result Question 40
44. The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass $M$ is also suspended, the period of oscillation will now be
[2010]
(a) $T$
(b) $T / \sqrt{2}$
(c) $2 T$
(d) $\sqrt{2} T$
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Answer:
Correct Answer: 44. (d)
Solution:
(d) $T=2 \pi \sqrt{\frac{M}{K}}$
$\therefore \quad \frac{T_1}{T_2}=\sqrt{\frac{M_1}{M_2}}$
$\therefore \quad T_2=T_1 \sqrt{\frac{M_2}{M_1}}=T_1 \sqrt{\frac{2 M}{M}}$
$T_2=T_1 \sqrt{2}=\sqrt{2} T(.$ where $.T_1=T)$
If the spring has a mass $M$ and mass $m$ is suspended from it, effective mass is given by $m _{\text{eff }}=m+\frac{M}{3}$
If the two mass $m_1$ and $m_2$ are connected by a spring and made to oscillate on horizontal surface, reduced mass $\frac{1}{m_r}=\frac{1}{m_1}+\frac{1}{m_2}$
$\therefore$ Time period $T=2 \pi \sqrt{\frac{m_r}{k}}$
