Question 66: Oscillations

Question

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?

Accepted Answer

From the potential energy curve,
$${U_{\max }} = {1 \over 2}k{A^2}$$
$$10 = {1 \over 2}k{(2)^2}$$
$$\Rightarrow$$ k = 5 N/m
The length of the simple pendulum, L = 4 m
Time period of spring,
$$T = 2\pi \sqrt {{k \over m}} $$
Time period of simple pendulum,
$$T = 2\pi \sqrt {{l \over g}} $$
The time period of simple pendulum is same as the time period of the spring oscillation.
$$\Rightarrow$$ $$2\pi \sqrt {{l \over g}} $$ = $$2\pi \sqrt {{k \over m}} $$
Substituting the values in the above equations, we get
$$2\pi \sqrt {{4 \over g}} $$ = $$2\pi \sqrt {{5 \over 5}} $$
g = 4m/s
2
$$	herefore$$ The acceleration due to the gravity on the planet is 4 m/s
2
.

Answer

10 m/s
2

Answer

5 m/s
2

Answer

4 m/s
2

Answer

9.8 m/s
2

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