Question

Two bodies of masses \(1\,kg\) and \(4\,kg\) are connected by a spring of spring constant \(K=5\,N\,m^{-1}\). Find the time period of oscillation.

• A. \(\frac{4\pi}{5}\,s\)
• B. \(\frac{2\pi}{5}\,s\)
• C. \(\pi\,s\)
• D. \(2\pi\,s\)

Hint:

For two masses connected by a spring, always use reduced mass \(\mu=\frac{m_1m_2}{m_1+m_2}\) before applying the SHM time-period formula.

Answer 1

The Correct Option is A

Concept: When two masses are connected by a spring and allowed to oscillate, the system performs simple harmonic motion. The effective mass of the system is the reduced mass. The time period is given by \[ T=2\pi\sqrt{\frac{\mu}{k}} \] where \[ \mu=\frac{m_1m_2}{m_1+m_2}. \]

Step 1: Calculate reduced mass. Given \[ m_1=1\,kg, \qquad m_2=4\,kg. \] Therefore \[ \mu=\frac{1\times4}{1+4} =\frac{4}{5}\,kg. \]

Step 2: Substitute in time period formula. Given \[ k=5\,N\,m^{-1}. \] Hence \[ T=2\pi\sqrt{\frac{4/5}{5}} \] \[ =2\pi\sqrt{\frac{4}{25}} \] \[ =2\pi\left(\frac{2}{5}\right) \] \[ =\frac{4\pi}{5}\,s. \]

Step 3: Final answer. \[ \boxed{T=\frac{4\pi}{5}\,s} \] Hence, \[ \boxed{(A)} \]