If the length of the oscillating simple pendulum is made 3 1 times the original keeping amplitude same then increase in its total energy at a place will be

Question

If the length of the oscillating simple pendulum is made 3 1 ​ times the original keeping amplitude same then increase in its total energy at a place will be

TestHub · Accepted Answer

The total energy of a simple pendulum is given by $E = \frac{1}{2} m ω^{2} A^{2}$ .

For a simple pendulum, the angular frequency squared is $ω^{2} = \frac{g}{L}$ .

Substituting this into the energy equation, we get:

$E = \frac{1}{2} m (\frac{g}{L}) A^{2}$

Since the mass $m$ , acceleration due to gravity $g$ , and amplitude $A$ are constant, the total energy is inversely proportional to the length of the pendulum:

$E \propto \frac{1}{L}$

When the length is made $\frac{1}{3}$ times the original length ( $L^{'} = \frac{L}{3}$ ), the new total energy $E^{'}$ becomes:

$E^{'} = \frac{1}{2} m (\frac{g}{L /3}) A^{2} = 3 (\frac{1}{2} m \frac{g}{L} A^{2}) = 3 E$

The question asks for the increase in total energy, which is:

$Δ E = E^{'} - E = 3 E - E = 2 E$

Thus, the increase in total energy is $2$ times the original energy.

Answer: 2 times

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