First slide

Simple hormonic motion

Question

A linear harmonic oscillator of force constant $2 \times 10^{6} {Nm}^{- 1}$ and amplitude 0.01 m has a total mechanical energy 160 J. Among the following statements, which are correct?
i. Maximum PE is 100 J
ii. Maximum KE is 100 J
iii. Maximum PE is 160 J
iv. Minimum PE is zero

Moderate

A

Both (i) and (iv)
B

Both (ii) and (iii)
C

Both (i) and (ii)
D

Both (ii) and (iv)

Solution

Total mechanical energy is $E_{T} = 160 J$
$∴ U_{\max} = 160 J$
At extreme position KE is zero. Work done by spring force from extreme position to mean position is
$W = \frac{1}{2} {kA}^{2}$
$∴ K_{\max} = W = \frac{1}{2} (2 \times 10^{6}) {(0.01)}^{2} = 100 J$
$∴ U_{\min} = 160 - 100 = 60 J$

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