Ncert Exercises & Solutions

In the given progressive wave $y = 5 \sin (100 πt - 0.4 πx)$ where y and x are in metre, t is in seconds. What is:

1. amplitude?

2. wavelength?

3. frequency?

4. wave velocity?

5. particle velocity amplitude?

Hint: Use the standard equation of the wave.

Step 1: Compare the given equation with the standard equation of the wave.

The standard equation of a progressive wave is given by:

y = asin (ωt - kx + ϕ)

This is travelling along positive x-direction.

Given equation is,

y = 5 \sin (100 πt - 0.4 πx)

Comparing with the standard equation;

1. Amplitude = 5m

k = \frac{2 π}{λ} = 0.4 π

∴ Wavelength, λ = \frac{2 π}{k} = \frac{2 π}{0.4 π} = \frac{20}{4} = 5 m

ω = 100 π

ω = 2 πν = 100 π

∴ Frequency, ν = \frac{100 π}{2 π} = 50 Hz

4. Wave velocity,

ν = \frac{ω}{k},

where k is wave number and

k = \frac{2 π}{λ}

v = \frac{100 π}{0.4 π} = \frac{1000}{4}

= 250 m / s

y = 5 \sin (100 πt - 0.4 πx)

\frac{dy}{dt} = particle velocity

.....(i)

From Eq. (i):

\frac{dy}{dt} = 5 (100 π) \cos [100 πt - 0.4 πx]

For particle, velocity amplitude=

{(\frac{dy}{dt})}_{m a x}

which will be for

{\{\cos [100 πt - 0.4 πx]\}}_{m a x} = 1

Particle velocity amplitude = {(\frac{dy}{dt})}_{\max} = 5 (100 π) \times 1

= 500 π m / s