Question

The average energy in one time period in simple harmonic motion is
(a) 12m ω2A2
(b) 14m ω2A2
(c) m ω2A2
(d) zero

Answer 1

The average energy in one time period in simple harmonic motion is $\frac{1}{2}mw^2a^2$

• The total energy of the Simple Harmonic Motion in One Time Period is called the average energy.  
• Energy depends only upon the magnitude but not on the direction, as it is a scalar quantity and not the vector quantity.
• Thus,
• the total energy of the Simple Harmonic Motion = Potential energy+Kinetic Energy  
• $=\dfrac{1}{2}kx^2 + \dfrac{1}{2}k(a^2-x^2)}\\\\=\dfrac{1}{2}kx^2 +\dfrac{1}{2}ka^2-\dfrac{1}{2}kx^2\\\\=\dfrac{1}{2}ka^2\\\\=\dfrac{1}{2}mw^2a^2$
• where,  
• a = amplitude
• w= angular frequency of the particle
• m = mass of the particle
• k = constant.

Answer 2

In simple harmonic motion, the average  energy consumed in one time period is $12 m \omega^{2} A^{2}$.  

Explanation:

Let us consider that at any instant, the particle has the displacement in SHM is    y = asin ωt … (1).

Here, ω is the particle’s angular frequency and a is the amplitude.

At any instant, the PE is  $PE = 1/2 ky^2 =\frac{1}{2} m \omega^2 y^2$

Where m is the particle’s mass and k is the constant.

The particle has the average potential energy over a whole period,  

 $1 T=\frac{1}{T} \int_{0}^{T} \frac{1}{2} m \omega^{2} y^{2} d t=$

 $\frac{1 m \omega^{2} a^{2}}{4}$         ...(iii)

At any instant, the particle has the velocity,$v=\frac{d y}{d t}=a \omega \cos \omega t$.

So, K.E. of the particle at any instance is

$=\frac{1}{T} \int_{0}^{T} \frac{1}{2} m v^{2} d t$

$=\frac{1}{T} \int_{0}^{T} \frac{1}{2} m a^{2} \omega^{2} \cos ^{2} \omega t d t=\frac{1}{4} m a^{2} \omega^{2}$

So, total energy average energy of the particle in SHM over a whole time period is = average P.E over a time period T+ average K.E over a time period T.

$\frac{1}{4} m a^{2} \omega^{2}+\frac{1}{4} m a^{2} \omega^{2}=\frac{1}{2} m a^{2} \omega^{2}$