The potential energy of a simple harmonic osc | vedclass

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Question

(A) Given: Total energy $E = 9 \,J$, Potential energy at mean position $U_{mean} = 5 \,J$, Mass $m = 2 \,kg$, Amplitude $A = 1 \,cm = 10^{-2} \,m$. Th

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$\frac{\pi}{100} \,s$

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(A) Given: Total energy $E = 9 \,J$, Potential energy at mean position $U_{mean} = 5 \,J$, Mass $m = 2 \,kg$, Amplitude $A = 1 \,cm = 10^{-2} \,m$. The total energy of an $SHM$ is the sum of kinetic and potential energy. At the mean position, the potential energy is $U_{mean} = 5 \,J$. Therefore, the maximum kinetic energy $K_{max}$ at the mean position is $K_{max} = E - U_{mean} = 9 \,J - 5 \,J = 4 \,J$. In $SHM$, the maximum kinetic energy is equal to the maximum potential energy at the extreme position, which is given by $\frac{1}{2} k A^2$. So, $\frac{1}{2} k A^2 = 4 \,J$. Substituting $A = 10^{-2} \,m$: $\frac{1}{2} k (10^{-2})^2 = 4 \implies \frac{1}{2} k (10^{-4}) = 4 \implies k = 8 	imes 10^4 \,N/m$. The time period $T$ is given by $T = 2 \pi \sqrt{\frac{m}{k}}$. Substituting the values: $T = 2 \pi \sqrt{\frac{2}{8 	imes 10^4}} = 2 \pi \sqrt{\frac{1}{4 	imes 10^4}} = 2 \pi 	imes \frac{1}{2 	imes 10^2} = \frac{\pi}{100} \,s$.

The potential energy of a simple harmonic oscillator of mass $2 \,kg$ at its mean position is $5 \,J$. If its total energy is $9 \,J$ and amplitude is $1 \,cm$, then its time period is

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