(A) The system consists of two masses $m_1 = 1 \ kg$ and $m_2 = 200 \ g = 0.2 \ kg$ connected by a spring of constant $k = 150 \ N/m$.For a two-body s

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(A) The system consists of two masses $m_1 = 1 \ kg$ and $m_2 = 200 \ g = 0.2 \ kg$ connected by a spring of constant $k = 150 \ N/m$.For a two-body spring-mass system, the equivalent mass (reduced mass) $\mu$ is given by:$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{1 \times 0.2}{1 + 0.2} = \frac{0.2}{1.2} = \frac{1}{6} \ kg$.The angular frequency $\omega$ of the system is given by:$\omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{150}{1/6}} = \sqrt{150 \times 6} = \sqrt{900} = 30 \ rad/s$.Thus, the angular frequency is $30 \ rad/s$.

Class 11 Physics Oscillations SHM of Spring Mass System

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As shown in the figure, a spring with spring constant $k = 150 \ N/m$ is kept in a stretched position by holding two masses of $1 \ kg$ and $200 \ g$ at a separation greater than the natural length of the spring. When the masses are released, assuming the horizontal surface to be frictionless, the angular frequency (in $SI$ units) of the system is:

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$30$
B
$27$
C
$20$
D
$5$

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