SATHEE: Oscillations - Result Question 40

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) $T / \sqrt{2}$

(d) $\sqrt{2} T$

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Answer:

Correct Answer: 44. (d)

Solution:

(d) $T = 2 π \sqrt{\frac{M}{K}}$

$∴ \frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{1}}{M_{2}}}$

$∴ T_{2} = T_{1} \sqrt{\frac{M_{2}}{M_{1}}} = T_{1} \sqrt{\frac{2 M}{M}}$

$T_{2} = T_{1} \sqrt{2} = \sqrt{2} T (.$ where $. T_{1} = T)$

If the spring has a mass $M$ and mass $m$ is suspended from it, effective mass is given by $m_{eff} = m + \frac{M}{3}$

If the two mass $m_{1}$ and $m_{2}$ are connected by a spring and made to oscillate on horizontal surface, reduced mass $\frac{1}{m_{r}} = \frac{1}{m_{1}} + \frac{1}{m_{2}}$

$∴$ Time period $T = 2 π \sqrt{\frac{m_{r}}{k}}$

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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

Answer:

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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