Question

If the length of a simple pendulum is doubled keeping its amplitude constant its energy will be-
A Unchanged
B Doubled
C Halved
D Quadrupled

Answer 1

The total energy of a particle performing simple harmonic motion is given as $E = \dfrac{1}{2}k{A^2}$. We express k in term of length using simple formulae of SHM to get the relation between energy and length i.e. $E \propto \dfrac{1}{l}$

Complete step by step answer:
We know that for a body performing simple harmonic oscillation is given by $E = \dfrac{1}{2}k{A^2}$, where $k = m{\omega ^2}$.m is the mass of pendulum and $\omega$ is the angular frequency.
Therefore, $E = \dfrac{1}{2}m{\omega ^2}{A^2}$. --------(1)
The angular frequency$\omega = \dfrac{{2\pi }}{T}$.
Squaring both sides we get,
$\Rightarrow {\omega ^2} = \dfrac{{4{\pi ^2}}}{{{T^2}}}$ ----------(2)
We know that time period of a simple pendulum is given by, $T = 2\pi \sqrt {\dfrac{l}{g}}$
Squaring both sides, ${T^2} = 4{\pi ^2}\dfrac{l}{g}$.
Substituting the value of T$^2$in equation (2) we get,
$\Rightarrow {\omega ^2} = \dfrac{{4{\pi ^2}}}{{4{\pi ^2}\dfrac{l}{g}}} = \dfrac{g}{l}$
Substituting the value of ${\omega ^2}$in equation (1) we get,
$\Rightarrow E = \dfrac{1}{2}m\dfrac{g}{l}{A^2}$.
Here m, g and A are constant. Therefore $E \propto \dfrac{1}{l}$.
Therefore if the length is doubled its energy becomes half.

Hence the correct option is C.

Additional information: Since for a given S.H.M., the mass of body m, angular speed ω and amplitude a are constant, Hence the total energy of a particle performing S.H.M. at C is constant i.e. the total energy of a linear harmonic oscillator is conserved. It is the same at all positions.
When the kinetic energy is maximum, the potential energy is zero. This occurs when the velocity is maximum and the mass is at the equilibrium position. The potential energy is maximum when the speed is zero. The total energy is the sum of the kinetic energy plus the potential energy and it is constant.

Note: The total energy for a particle performing SHM is given as$E = \dfrac{1}{2}k{A^2}$, and the time period for a simple pendulum is $T = 2\pi \sqrt {\dfrac{l}{g}}$.We should know the simple formula of particle performing SHM to solve such type of problems.