Question

Find the distance between two successive nodes in a stationary wave on a string vibrating with frequency 64Hz. The velocity of progressive wave that resulted in the stationary wave is 48 m s^-1.​

Answer 1

♦Given:-

$\: \: \: \: \: \: \: \: \: \: \: \: \: •. Speed of wave = v = 48 Ms</p><p> \\ • Frequency n = 64 Hz</p><p>$

♦ We have,

$v = n \: λ$

$=> λ= \: \: \frac {v} {n} \\ \\ = \frac{48}{64} \\ \\ = 0.75 \: m$

*We know that distance between successive nodes

$\: \: \: = \frac{λ}{2} \\ \\ = \frac{0.75}{2} \\ \\ = 0.375 \: m$

Answer 2

Given:

The velocity of progressive wave that resulted in the stationary wave is 48 m s^-1 and frequency is 64 Hz.

To find:

Distance between two consecutive nodes?

Calculation:

• Let wavelength of wave be $\lambda$.

$\therefore \lambda = \dfrac{v}{f}$

$\implies\lambda = \dfrac{48}{64}$

$\implies\lambda = 0.75 \: m$

Now, we know that:

• Distance between consecutive nodes is $\lambda/2$.

• Let required distance be d :

$\implies d = \dfrac{\lambda}{2} = 0.375 \: m$

So, distance between consecutive nodes is 0.375 metres.