Question

In the given figure, a mass \(M\) is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is \(k\). The mass oscillates on a frictionless surface with the time period \(T\) and amplitude \(A\). When the mass is in an equilibrium position, as shown in the figure, another mass \(m\) is gently fixed upon it. The new amplitude of oscillation will be : 

• A. $A \sqrt{\frac{ M - m }{ M }}$
• B. $A \sqrt{\frac{ M }{ M + m }}$
• C. $A \sqrt{\frac{ M + m }{ M }}$
• D. $A \sqrt{\frac{M}{M-m}}$

Answer 1

The Correct Option is B

To determine the new amplitude of oscillation when another mass \( m \) is gently placed on the mass \( M \) attached to the spring, we can use the principles of conservation of mechanical energy and properties of simple harmonic motion (SHM).

Initially, the energy in the system when only mass \( M \) is oscillating is given by:

E_i = \frac{1}{2} k A^2

After placing the additional mass \( m \) onto \( M \) at equilibrium, the system's total mass becomes \( M + m \). The new mechanical energy ( \( E_f \) ) will be distributed in the oscillation of the combined mass.

Since there is no external work done and assuming no energy loss:

E_i = E_f

The energy in the system with the new combined mass and the new amplitude \( A' \) is:

E_f = \frac{1}{2} k A'^2

Setting the initial and final energy equal gives us:

\frac{1}{2} k A^2 = \frac{1}{2} k A'^2

Since the energy before and after must remain the same:

\frac{1}{2} k A^2 = \frac{1}{2} k A'^2

However, with the new mass, we must consider the change in system properties. Using conservation of momentum principles, the kinetic energy at the equilibrium translates proportionally:

A' = A \sqrt{\frac{M}{M + m}}

Thus, the correct option is:

$A \sqrt{\frac{ M }{ M + m }}$

This solution illustrates the effect of adding an additional mass on the oscillating system's amplitude, accounting for unchanged energy and altered system mass.