Question

Find the distance between two successive antinodes in a stationary wave on a string vibrating with frequency 32 Hz. [ Speed of wave = 48 m/s. ]​

Answer 1

Answer:

0.75m

Explanation:

Let wavelength = d
Then length between two successive antinodes will be d/2.

frequency = 32Hz

velocity = 48m/s

v = fd

d = v/f

d = 48/32 = 3/2 m = wavelength

length between two successive antinodes = d/2 = 3/4 = 0.75m

Answer 2

DISTANCE BETWEEN SUCCESSIVE ANTI NODES IS 0.75 METRES.

Given:

• Velocity of wave = 48 m/s.
• Frequency = 32 Hz

To find:

• Distance successive antinodes in the stationary wave?

Calculation:

First of all, let's calculate the wavelength of the stationary wave :

$v = f \times \lambda$

$\implies \: \lambda = \dfrac{v}{f}$

$\implies \: \lambda = \dfrac{48}{32}$

$\implies \: \lambda = 1.5 \: metre$

• Now, we know that the distance between two successive antinodes in a stationary wave is $\lambda/2$

So, let the distance be d :

$\implies \: d = \dfrac{ \lambda}{2}$

$\implies \: d = \dfrac{1.5}{2}$

$\implies \: d = 0.75 \: m$

So, is between successive antinode is 0.75 metres.