Question
Derive an expression for the potential energy of an elastic stretched spring.
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Answer
Consider a spring attached with a mass m stretching by a length y after t sec. Restoring force F = mass × acceleration$=-\text{}\omega^2\text{y}=-\text{ky}$
where, $\text{k = spring constant = m}\omega^2$ Work done for an additional displacement dy against restoring force is$\text{dW}=-\text{F dy}$
$=-(-\text{ky})\text{dy = ky dy}$
Total work done$\text{W}=\int\limits^{\text{y}}_0\text{ky dy}=\frac{1}{2}\text{ky}^2$
This work done appears as a PE. 'U' of the particle.$\text{U}=\frac{1}2{}\text{ky}^2=\frac{1}{2}\text{m}\omega^2\text{y}^2$
$=\frac{1}{2}\text{m}\omega^2\text{a}^2\sin^2\omega\text{t}$
where, $\text{k = spring constant = m}\omega^2$ Work done for an additional displacement dy against restoring force is$\text{dW}=-\text{F dy}$
$=-(-\text{ky})\text{dy = ky dy}$
Total work done$\text{W}=\int\limits^{\text{y}}_0\text{ky dy}=\frac{1}{2}\text{ky}^2$
This work done appears as a PE. 'U' of the particle.$\text{U}=\frac{1}2{}\text{ky}^2=\frac{1}{2}\text{m}\omega^2\text{y}^2$
$=\frac{1}{2}\text{m}\omega^2\text{a}^2\sin^2\omega\text{t}$
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