JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 5)
An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $$t = {T \over 4}s$$ starting from mean position. Assume that the initial phase of the oscillation is zero.
0.62 J
6.2 $$\times$$ 10$$-$$3 J
1.2 $$\times$$ 103 J
6.2 $$\times$$ 103 J
Explanation
$$T = 2\pi \sqrt {{m \over k}} $$
$$0.2 = 2\pi \sqrt {{{0.5} \over k}} $$
k = 50$$\pi$$2
$$ \approx $$ 500
x = A sin ($$\omega$$t + $$\phi$$)
= 5 cm sin $$\left( {{{\omega T} \over 4} + 0} \right)$$
= 5 cm sin $$\left( {{\pi \over 2}} \right)$$
= 5 cm
$$PE = {1 \over 2}k{x^2}$$
$$ = {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}$$
= 0.6255
$$0.2 = 2\pi \sqrt {{{0.5} \over k}} $$
k = 50$$\pi$$2
$$ \approx $$ 500
x = A sin ($$\omega$$t + $$\phi$$)
= 5 cm sin $$\left( {{{\omega T} \over 4} + 0} \right)$$
= 5 cm sin $$\left( {{\pi \over 2}} \right)$$
= 5 cm
$$PE = {1 \over 2}k{x^2}$$
$$ = {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}$$
= 0.6255
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