Two masses m1=1 kg and m2=0.5 kg are suspended together by a massless spring of spring constant 12.5 Nm -1. When masses are in equilibrium m1 is removed without disturbing the system. New amplitude of oscillation will be

Question

Two masses m1=1 kg and m2=0.5 kg are suspended together by a massless spring of spring constant 12.5 Nm -1. When masses are in equilibrium m1 is removed without disturbing the system. New amplitude of oscillation will be (A) 30 cm (B) 50 cm (C) 80 cm (D) 60 cm

Tardigrade Admin · Accepted Answer

Points of equilibrium of the spring will be when no force acts on it.
or

k x = (m_{1} + m_{2}) g

x = \frac{( m _{1} + m _{2} ) g}{k}

The new equilibrium position which will be the mean position of S.H.M. will be simply

\frac{m _{2} g}{k}

New amplitude will be maximum displacement from

\frac{m _{2} g}{k}

which is :

A = \frac{( m _{1} + m _{2} ) g}{k} - \frac{m _{2} g}{k}

or

A = \frac{m _{1} g}{k}

or

A = \frac{1 \times 10}{12.5}

or

A = \frac{4}{5} m

∴ A = 0.8 m

or

80 c m

Q. Two masses m1​=1kg and m2​=0.5kg are suspended together by a massless spring of spring constant 12.5Nm−1. When masses are in equilibrium m1​ is removed without disturbing the system. New amplitude of oscillation will be

30 cm

50 cm

80 cm

60 cm

Q. Two masses $m_{1} = 1 k g$ and $m_{2} = 0.5 k g$ are suspended together by a massless spring of spring constant $12.5 N m^{- 1}$ . When masses are in equilibrium $m_{1}$ is removed without disturbing the system. New amplitude of oscillation will be