In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is :

Solution

Physics sir · 3 steps

Gravity only shifts where the block rests; the two springs decide how fast it oscillates.

1. Springs in parallel. Push the block a distance x up the incline. The lower spring stretches by x and the upper spring compresses by x, so both pull it back down the slope. The restoring force is kx + kx = 2kx, so k_(eff) = 2k.
2. Gravity drops out. The component Mg sin α is constant. It only moves the equilibrium point slightly down the slope. Measure x from that new equilibrium and the net force is still -2kx, with no g or α left in it.

   f = 1/2π√(k_(eff)/M) = 1/2π√(2k/M)

3. So. The frequency is 1/2π√(2k/M), which is option (c).

The idea to keep. A constant force like Mg sin α never changes the frequency of a spring system. It only shifts the equilibrium. Find k_(eff) and use f = 1/2π√(k_(eff)/M).

Answer from the NTA final answer key. Solution by PrepWiser.