Potential Energy In Simple Harmonic Motion

Potential Energy

If we analyze the whole motion of the spring-mass system we can figure out there are two types of energy. One is due to the motion of the block or the kinetic energy while the other is due to the spring known as elastic potential energy. During the motion of the block, the energy transformation takes place from kinetic to potential energy and vice versa. Now that we know the types of energy we will try to find them as a function of time.

For SHM,

\(x (t)\) = \(A sin (ωt)\)

Where,\(x (t)\) is the displacement from the mean position or the normal length of the spring

Potential Energy of the spring = \(\frac{kx^2}{2}\)

\(P.E\) = \(\frac{1}{2}kA^2 sin^2(ωt)\)

Now the velocity of the block = \(Aω cos (ωt)\)

\(K.E\) = \(\frac{1}{2} mv^2\)

\(K.E\) = \(\frac{1}{2}mA^{2}ω^2 cos^{2}(ωt)\)

Also, \(mω^2\) = \(k\)

Substituting this we get,

\(K.E\) = \(\frac{1}{2} kA^2cos^2(ωt)\)

From the above two equation, we can figure out that total energy is constant.

\(K.E + P.E\) = \(\frac{1}{2} kA^2 [ sin^2(ωt) + cos^2(ωt)]\)

\(K.E + P.E\) = \(\frac{1}{2}kA^2\)

This is the total mechanical energy of the system. So now that we have mathematical formulas we try to analyze the energies using graphs.

From the graph, we infer that potential energy is maximum at extreme position and minimum at the mean position.

Similarly, the kinetic energy is maximum at mean position and minimum at the extreme positions.

Sample Problem: Find the displacement when potential energy of the system is equal to kinetic energy.

\(K.E\) = \(\frac{1}{2} kA^2 cos^2(ωt)\)

\(P.E\) = \(\frac{1}{2}kA^2 sin2(ωt)\)

Equating the two we get,

\(cos^2(ωt) – sin^2(ωt)\) = \(0\)

\(cos (2 ωt)\) = \(0\)

Therefore, \(2 ωt\) = \(\frac{π}{2}\)

\(ωt\) = \(\frac{π}{4}\)

Putting this value, we get,

\(x\) = \(\frac{A}{√2}\)

So at this displacement, potential energy becomes equal to kinetic energy.

If the time period of oscillation is ‘T’ what will be the time period of the maximum difference between potential energy and kinetic energy? To explore more on physics calculators join us here at BYJU’S

Practise This Question

A series RLC circuit contains resistance of 10Ω, an inductance of 0.2H and a capacitor of 100uF .. Its powered by a supply of 200V, 50Hz. Calculate the total circuit impedance, the circuits current, power factor and draw the voltage phasor diagram.