Ncert Exercises & Solutions

A body is performing SHM, then its:

(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.

(d) root mean square velocity is $\frac{1}{\sqrt{2}}$ times of its maximum velocity.

Choose the correct alternatives:

1. (a, b, d)

2. (a, c)

3. (b, d)

4. (b, c, d)

(1) Hint: Use the formula of kinetic energy and the total energy of SHM.

Let the equation of an SHM is represented as x=asin

ω t

Assume the mass of the body is m.

(a) Step 1: Find the maximum kinetic energy and average total energy of the particle.

The total mechanical energy of the body at any time t is,

E = \frac{1}{2} m ω^{2} a^{2}

...(i)

Kinetic energy at any instant t is,

K = \frac{1}{2} m v^{2} = \frac{1}{2} m {[\frac{d x}{d t}]}^{2} [∵ v = \frac{d x}{d t}]

= \frac{1}{2} m ω^{2} a^{2} \cos^{2} ω t

\Rightarrow K_{m a x} = \frac{1}{2} m ω^{2} a^{2} = E [∵ f o r k_{m a x}, \cos ω t = 1] . . . (i i)

(b) Step 2: Find the average kinetic energy of the particle.

KE at any instant t is,

K = \frac{1}{2} m ω^{2} a^{2} \cos^{2} ω t

(K_{a v})

for a cycle

= \frac{1}{2} m ω^{2} a^{2} {(\cos^{2} ω t)}_{a v}

for a cycle

= \frac{1}{2} m ω^{2} a^{2} [\frac{0 + 1}{2}]

= \frac{1}{4} m ω^{2} a^{2} = \frac{K_{m a x}}{2}

[from Eq.(ii)]

Step 3: Find the mean velocity and the RMS velocity of the particle.

\frac{d x}{d t} = a ω \cos ω t

v_{m e a n} = \frac{v_{m a x} + v_{m i n}}{2}

= \frac{a ω + (- a ω)}{2} = 0

[For a complete cycle]

v_{m a x} \neq v_{m e a n}

(d)

v_{r m s} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2}}{2}} = \sqrt{\frac{0 + a^{2} ω^{2}}{2}} = \frac{a ω}{\sqrt{2}}

\Rightarrow v_{r m s} = \frac{v_{m a x}}{\sqrt{2}}