Question

The period of oscillation of a mass \(M\) suspended from a spring of negligible mass is \(T\). If along with it another mass \(M\) is also suspended, the period of oscillation now will be

• A. \(T\)
• B. \(2T\)
• C. \(\dfrac{T}{\sqrt{2}}\)
• D. \(\sqrt{2}\,T\)

Hint:

In a spring-mass system, time period varies as the square root of the attached mass.

Answer 1

The Correct Option is D

Step 1: Write the formula for time period of a spring-mass system.
The time period of oscillation is given by:
\[ T = 2\pi \sqrt{\frac{M}{k}} \] where \(M\) is the mass and \(k\) is the spring constant.

Step 2: New mass after adding another mass.
When another mass \(M\) is suspended, the total mass becomes:
\[ M' = 2M \]
Step 3: Write the new time period.
\[ T' = 2\pi \sqrt{\frac{2M}{k}} \]
Step 4: Express the new period in terms of \(T\).
\[ T' = \sqrt{2}\left(2\pi \sqrt{\frac{M}{k}}\right) \] \[ T' = \sqrt{2}\,T \]
Step 5: Conclusion.
The new time period of oscillation is \(\sqrt{2}\,T\).