A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed?

Solution

Physics sir · 4 steps

Read k off the energy curve, then match the pendulum's ω² = g/L to the spring's k/m.

1. Find the centre. The curve has its minimum at x = 2 m, so that is the equilibrium position. Displacement is measured from there, not from the origin.
2. Find k. At x = 0 and x = 4 m the displacement is 2 m and U = 10 J. Using U = 1/2k(x-2)² gives 10 = 1/2k × 4, so k = 5 N m⁻¹.

   k/m = g/L ⇒ g = kL/m

3. Equal periods. Equal periods mean equal ω². With k/m = 5/5 = 1 s⁻², we get g = 1 × 4 = 4 m s⁻².
4. So. The acceleration due to gravity on this planet is 4 m s⁻², option (c).

The idea to keep. On a U–x curve, the minimum is the equilibrium point. Measure displacement from there, use U = 1/2kx² to get k, and equal periods means k/m = g/L.

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