JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 15)
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?
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10 m/s2
5 m/s2
4 m/s2
9.8 m/s2
Explanation
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/s2
$$\therefore$$ The acceleration due to the gravity on the planet is 4 m/s2.
$${U_{\max }} = {1 \over 2}k{A^2}$$
$$10 = {1 \over 2}k{(2)^2}$$
$$\Rightarrow$$ 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/s2
$$\therefore$$ The acceleration due to the gravity on the planet is 4 m/s2.
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