Question 46: Oscillations

Question

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

Accepted Answer

For a two-body system, the angular frequency ( $\omega$ ) is given by :
$$ \omega=\sqrt{\frac{\mathrm{k}}{\mu}} $$
Where :
k is the spring constant.
$\mu$ is the reduced mass of the system.
The reduced mass is calculated using the formula :
$$ \mu=\frac{m_1 \cdot m_2}{m_1+m_2} $$
Given values of masses are :
$\mathrm{m}_1=1 \mathrm{~kg}$
$\mathrm{m}_2=200 \mathrm{~g}=0.2 \mathrm{~kg}$
$$ \mu=\frac{1 	imes 0.2}{1+0.2}=\frac{0.2}{1.2}=\frac{1}{6} \mathrm{~kg} $$
So the angular frequency is,
$$ \omega=\sqrt{\frac{150}{1 / 6}} $$
$$\Rightarrow $$ $$ \omega=\sqrt{900}=30 \mathrm{rad} / \mathrm{s} $$
Therefore, the angular frequency of the system in SI units is 30 . Hence, the correct option is (C).

Answer

20

Answer

5

Answer

30

Answer

27

Oscillations

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