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Question 29

As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 1 kg and 0.2 kg with a separation more than spring natural length and are released. Assuming the horizontal serface to be frictionless, the angular frequency (in SI unit) of the system is:

image

This is a two-mass spring system, so oscillation occurs with reduced mass.

For two masses connected by a spring,

$$\omega=\sqrt{\frac{k}{\mu}}$$

where reduced mass

$$\mu=\frac{m_1m_2}{m_1+m_2}$$

Given

$$m_1=1kg$$

$$m_2=0.2kg$$

So

$$\mu=\frac{(1)(0.2)}{1+0.2}$$

$$=\frac{0.2}{1.2}$$

$$=\frac{1}{6}$$

Now

$$\omega=\sqrt{\frac{150}{1/6}}$$

$$=\sqrt{900}$$

$$=30\ \text{rad/s}$$

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