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NCERT Exemplar · Q17

Q.A body is performing S.H.M. Then its (Note: more than one of the given options may be correct.)

(a) average total energy per cycle is equal to its maximum kinetic energy.
(b) average kinetic energy per cycle is equal to half of its maximum kinetic energy.
(c) mean velocity over a complete cycle is equal to 2π\dfrac{2}{\pi} times of its maximum velocity.
(d) root mean square velocity is 12\dfrac{1}{\sqrt{2}} times of its maximum velocity.
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In Simple Harmonic Motion (SHM), the total energy remains constant, so its average over a cycle equals its maximum kinetic energy. The average kinetic energy and the root mean square (RMS) velocity over a cycle are both related to half of their maximum values, while the mean velocity over a cycle is zero. Options (A), (B), and (D) are correct.

Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction. This leads to oscillatory motion described by sinusoidal functions. Understanding the time-varying nature of displacement, velocity, and energy is crucial for analyzing their average values over a complete cycle.

Let's consider a body performing SHM. We can describe its displacement from the equilibrium position as:

x(t)=Asin⁡(ωt)x(t) = A \sin(\omega t)

where AA is the amplitude, ω\omega is the angular frequency, and tt is time. The period of oscillation is T=2πωT = \frac{2\pi}{\omega}.

From this, we can derive the velocity and acceleration:

Velocity: v(t)=dxdt=Aωcos⁡(ωt)v(t) = \frac{dx}{dt} = A\omega \cos(\omega t)

Acceleration: a(t)=dvdt=−Aω2sin⁡(ωt)=−ω2x(t)a(t) = \frac{dv}{dt} = -A\omega^2 \sin(\omega t) = -\omega^2 x(t)

The maximum velocity occurs when cos⁡(ωt)=±1\cos(\omega t) = \pm 1, so vmax=Aωv_{max} = A\omega.

Now, let's analyze each option.

1. Option (A): Average total energy per cycle is equal to its maximum kinetic energy.

The total mechanical energy (EE) in SHM is the sum of its kinetic energy (KK) and potential energy (UU).

K(t)=12mv(t)2=12m(Aωcos⁡(ωt))2=12mA2ω2cos⁡2(ωt)K(t) = \frac{1}{2} m v(t)^2 = \frac{1}{2} m (A\omega \cos(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)

U(t)=12kx(t)2=12mω2(Asin⁡(ωt))2=12mA2ω2sin⁡2(ωt)U(t) = \frac{1}{2} k x(t)^2 = \frac{1}{2} m \omega^2 (A \sin(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t) (since k=mω2k = m\omega^2)

The total energy is:

E=K(t)+U(t)=12mA2ω2(cos⁡2(ωt)+sin⁡2(ωt))E = K(t) + U(t) = \frac{1}{2} m A^2 \omega^2 (\cos^2(\omega t) + \sin^2(\omega t))

Since cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1,

E=12mA2ω2E = \frac{1}{2} m A^2 \omega^2

Important

The total mechanical energy in SHM is constant and does not vary with time.

The maximum kinetic energy (KmaxK_{max}) occurs when the velocity is maximum (vmax=Aωv_{max} = A\omega), i.e., at the equilibrium position (x=0x=0).

Kmax=12mvmax2=12m(Aω)2=12mA2ω2K_{max} = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m (A\omega)^2 = \frac{1}{2} m A^2 \omega^2

Since the total energy EE is constant and equal to KmaxK_{max}, its average value over any time interval (including a complete cycle) will simply be EE.

Therefore, the average total energy per cycle is equal to its maximum kinetic energy.

Option (A) is correct.

2. Option (B): Average kinetic energy per cycle is equal to half of its maximum kinetic energy.

We have the kinetic energy as K(t)=Kmaxcos⁡2(ωt)K(t) = K_{max} \cos^2(\omega t).

To find the average kinetic energy over a complete cycle (TT), we integrate K(t)K(t) over one period and divide by TT:

⟨K⟩=1T∫0TK(t)dt=1T∫0TKmaxcos⁡2(ωt)dt\langle K \rangle = \frac{1}{T} \int_0^T K(t) dt = \frac{1}{T} \int_0^T K_{max} \cos^2(\omega t) dt

⟨K⟩=KmaxT∫0Tcos⁡2(ωt)dt\langle K \rangle = \frac{K_{max}}{T} \int_0^T \cos^2(\omega t) dt

We use the trigonometric identity: cos⁡2θ=1+cos⁡(2θ)2\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}

⟨K⟩=KmaxT∫0T1+cos⁡(2ωt)2dt\langle K \rangle = \frac{K_{max}}{T} \int_0^T \frac{1 + \cos(2\omega t)}{2} dt

⟨K⟩=Kmax2T[t+sin⁡(2ωt)2ω]0T\langle K \rangle = \frac{K_{max}}{2T} \left[ t + \frac{\sin(2\omega t)}{2\omega} \right]_0^T

⟨K⟩=Kmax2T[(T+sin⁡(2ωT)2ω)−(0+sin⁡(0)2ω)]\langle K \rangle = \frac{K_{max}}{2T} \left[ (T + \frac{\sin(2\omega T)}{2\omega}) - (0 + \frac{\sin(0)}{2\omega}) \right]

Since T=2πωT = \frac{2\pi}{\omega}, we have 2ωT=2ω(2πω)=4π2\omega T = 2\omega \left(\frac{2\pi}{\omega}\right) = 4\pi.

So, sin⁡(2ωT)=sin⁡(4π)=0\sin(2\omega T) = \sin(4\pi) = 0.

⟨K⟩=Kmax2T[T+0−0]=Kmax2\langle K \rangle = \frac{K_{max}}{2T} [T + 0 - 0] = \frac{K_{max}}{2}

Thus, the average kinetic energy per cycle is half of its maximum kinetic energy.

Option (B) is correct.

Tip

By symmetry, the average potential energy over a cycle is also 12Umax\frac{1}{2} U_{max}. Since Umax=KmaxU_{max} = K_{max}, the average potential energy is also 12Kmax\frac{1}{2} K_{max}. This means ⟨K⟩=⟨U⟩=12E\langle K \rangle = \langle U \rangle = \frac{1}{2} E.

3. Option (C): Mean velocity over a complete cycle is equal to 2π\dfrac{2}{\pi} times of its maximum velocity.

The velocity of the body is v(t)=Aωcos⁡(ωt)v(t) = A\omega \cos(\omega t).

The maximum velocity is vmax=Aωv_{max} = A\omega. …

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