Question

A string of length $1m$and mass$5g$ is fixed at both ends. The tension in the string is $8.0N$. The siring is set into vibration using an external vibrator of frequency $100Hz$. The separation between successive nodes on the string is close to __?

• A. $10cm$
• B. $16.6cm$
• C. $20cm$
• D. $33.3cm$

Answer 1

The correct option is C

$20cm$

Step 1: Given data

Length of the string, $l=1m$

Mass of the string, $m=5g$

Mass per unit length, $\mu =5/1g/m$

Tension in the string, $T=8.0N$

Frequency of the wave, $n=100Hz$

Step 2: Find the velocity of wave on string, V

The velocity of wave on string,$V=\sqrt{(T/\mu )}$ [Where, $\mu =$mass per unit length]

$$\begin{gathered} =\surd (8\times 1000/5) \\ =40m/s \end{gathered}$$

Step 3: Find the separation between successive nodes

The wavelength of wave,$\lambda =v/n$

$=40/100m$

Separation between successive nodes, $\lambda /2=20/100m$

$=20cm$

Hence, option (C) is correct.