ENERGY IN SIMPLE HARMONIC MOTION -

When the particle executes S.H.M, it possesses both potential energy and kinetic energy. P.E is due to conservative force and K.E is due to its velocity.

Let the particle execute S.H.M as

$\fn_cm \large y=A\sin(\omega t+\phi)$

The P.E of the particle at time t is

$\fn_cm \large U=\frac{1}{2}ky^2$

$\fn_cm \large \left [ U=\frac{1}{2}kA^2 \sin^2(\omega t+\phi) \right ]$

The velocity of the particle at time t is given by

$\fn_cm \large v=A\omega\cos(\omega t+\phi))$

i.e its K.E at time t is given by

$\fn_cm \large K=\frac{1}{2}mv^2$

$\fn_cm \large K=\frac{1}{2}mA^2\omega^2\cos^2(\omega t+\phi)$

$\fn_cm \large \left [ K=\frac{1}{2}K A^2\cos^2(\omega t+\phi) \right ]\;\;\;\;\;(\because \frac{k}{m}=\omega^2)$

$\fn_cm \large \therefore$ Total energy of particle executing S.H.M at time t is

E=P.E+K.E

$\fn_cm \large E=\frac{1}{2}KA^2\sin^2(\omega t+\phi)+\frac{1}{2}KA^2\cos^2(\omega t+\phi))$

$\fn_cm \large E=\frac{1}{2}KA^2[\sin^2(\omega t+\phi)+\cos^2(\omega t+\phi)]$

$\fn_cm \large \left [ E=\frac{1}{2}KA^2 \right ]\;\;\;\;\;which\;is\;constant$

i.e when the particle executing S.H.M, the total energy must be constant. It shows that energy is conserved when the particle in S.H.M.

NOTE

1. Both K.E and P.E in S.H.M seem to be positive.

2. from K.E relation $\fn_cm \large K=\frac{1}{2}m\omega^2A^2sin^2(\omega t+\phi)=\frac{1}{2}m\omega^2A^2\left [ 1-cos2(\omega t+\phi) \right ]$

This function is also periodic in nature with angular frequency 2ω and time period T/2, but this is not S.H.M in nature because

$\fn_cm \large \frac{d^2K}{dt^2}\not{\propto} K$

Same as P.E also periodic in nature with angular frequency 2ω and time period T/2.

3. From the above relation $\fn_cm \large \left [ E=\frac{1}{2}KA^2 \right ]$ total energy of a particle executing SHM is directly proportional to the square of the amplitude. Now $\fn_cm \large E=\frac{1}{2}m\omega^2A^2=\frac{1}{2}m4 \pi^2f^2A^2$ i.e it is also proportional to the square of the frequency and mass of the particle.

IN SHM THE AVERAGE K.E OVER A PERIOD IS EQUAL TO THE AVERAGE P.E OVER THE SAME PERIOD

Consider a particle of mass m executing S.H.M with amplitude A and angular frequency ω. Then displacement Y of the particle from the mean position at any time t is given by:

$\fn_cm \large y=A\sin\omega t$

$\fn_cm \large \therefore velocity,\;\;v=\frac{d y}{dt}=A\omega\cos \omega t$

$\fn_cm \large K.E=\frac{1}{2}mv^2=\frac{1}{2}mA^2\omega^2\cos^2\omega t$

$\fn_cm \large P.E=\frac{1}{2}ky^2=\frac{1}{2}m\omega^2A^2\sin^2\omega t\;\;\;\;\;(\because k=m\omega^2)$

The average kinetic energy over one cycle is given by

$\fn_cm \large K.E_{avg}=\frac{1}{T}\int_{0}^{T}(K.E)dt=\frac{1}{T}\int_{0}^{T}\frac{1}{2}mA^2\omega^2\cos^2\omega tdt$

$\fn_cm \large =\frac{1}{2T}mA^2\omega^2\int_{0}^{T}\frac{1+cos2\omega t}{2}dt$

$\fn_cm \large =\frac{1}{4T}mA^2\omega^2\left [ t+\frac{sin2\omega t}{2\omega} \right ]_0^T=\frac{1}{4T}mA^2\omega^2(T)$

$\fn_cm \large K.E_{avg}=\frac{1}{4}mA^2\omega^2\;\;\;\;\;----------(I)$

The average potential energy over one cycle is given by

$\fn_cm \large P.E_{avg}=\frac{1}{T}\int_{0}^{T}(P.E)dt=\frac{1}{T}\int_{0}^{T}\frac{1}{2}m\omega^2A^2sin^2\omega tdt$

$\fn_cm \large =\frac{1}{2T}mA^2\omega^2\int_{0}^{T}\frac{1-cos2\omega t}{2}dt$

$\fn_cm \large =\frac{1}{4T}mA^2\omega^2\left [ t-\frac{sin2\omega t}{2\omega} \right ]_0^T=\frac{1}{4T}mA^2\omega^2(T)$

$\fn_cm \large P.E_{avg}=\frac{1}{4}mA^2\omega^2\;\;\;\;\;----------(II)$

From (I) and (II) we can say that average K.E and average P.E are the same over a period.

Q.1> A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N /m. The block is pulled to a distance x =10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.

Q.2> A mass of 0.5 kg connected to a light spring of force constant 20N/m oscillate on a horizontal frictionless table.

(a) Calculate the total energy of the system and the maximum speed of the mass if their amplitude of the motion is 3 cm.

(b) What is the velocity of the mass when the displacement is equal to 2 cm?

Q.3> A particle executes SHM of amplitude A. At what distance from the mean position its K.E is equal to its P.E?

Q.4> Two joules are required to extend a spring by 0.25 m. What is the spring constant?