A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become nA. The value of n is :

Complete Solution:

In SHM, the total mechanical energy E of a mass-spring system is conserved if there is no friction. The energy is given by:

E = ½ k A²

where k is the spring constant, and A is the amplitude.

At the equilibrium position, the entire energy of the system is kinetic. The kinetic energy for the block of mass m at the equilibrium position (where velocity is maximum) is:

E = ½ m v_max²

Since E = ½ k A², we can equate:

½ m v_max² = ½ k A²

Simplifying, we get:

v_max = A √(k/m)

When the block reaches the equilibrium point, half of its mass breaks off, reducing the mass from m to m/2. At this moment, since there is no external force acting on the system and it’s at equilibrium, the velocity v_max remains the same for the remaining mass m/2.

For the new mass m' = m/2, the mechanical energy of the system can be expressed as:

E' = ½ m' v_max²

Substitute m' = m/2:

E' = ½ · (m/2) · v_max² = (¼) m v_max²

We know m v_max² = k A², so:

E' = ¼ k A²

Since E' = ½ k (nA)² (where nA is the new amplitude), we equate:

½ k (nA)² = ¼ k A²

Dividing both sides by ½ k:

(nA)² = A² / 2

Taking the square root of both sides:

nA = A / √2

Therefore, n = 1/√2 or approximately 0.707.

Final Answer:

The value of n is 1/√2 or approximately 0.707.