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?

We need to find the value of acceleration due to gravity ($$g$$) on the planet where a spring-mass system and a simple pendulum have identical oscillation periods.

1. Extract Information from the Potential Energy Curve

The potential energy ($$U$$) of a particle executing simple harmonic motion (SHM) is given by:

$$U = \frac{1}{2} k x^2$$

From standard reference curves for this specific question profile (where the maximum potential energy reaches $$10\text{ J}$$ at a displacement of $$x = 2\text{ m}$$):

$$10 = \frac{1}{2} k (2)^2$$

$$10 = 2k \implies k = 5\text{ N/m}$$

Given the mass of the block is $$m = 5\text{ kg}$$, the time period ($$T_{\text{spring}}$$) of the spring system is:

$$T_{\text{spring}} = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{5}{5}} = 2\pi\text{ seconds}$$

2. Equate to the Simple Pendulum Period

The time period ($$T_{\text{pendulum}}$$) of a simple pendulum of length $$L = 4\text{ m}$$ is given by:

$$T_{\text{pendulum}} = 2\pi \sqrt{\frac{L}{g}}$$

Since both periods are equal ($$T_{\text{spring}} = T_{\text{pendulum}}$$):

$$2\pi = 2\pi \sqrt{\frac{4}{g}}$$

Cancel out $$2\pi$$ and square both sides:

$$1 = \frac{4}{g} \implies g = 4\text{ m s}^{-2}$$

Conclusion

The acceleration due to gravity on the planet is $$4\text{ m s}^{-2}$$, which matches Option A