Question

A vibrator makes 150 cm of a string to vibrate in 6 loops in the longitudinal arrangement when it is stretched by 150 N. The entire length of the string is then weighed and is found to weigh 400 mg. Then
A. Frequency of the vibrator is 3 kHz.
B. Frequency of the vibrator is 1.5 kHz.
C. Distance between two nodes is 25 cm.
D. Distance between two nodes is 33 cm.

Answer 1

Calculate the velocity of the wave produced on the string. Then determine the wavelength of the wave using the condition given in the question. Use the relation between frequency, wavelength and velocity to determine the frequency of the wave. We know that the distance between the two nodes is equal to half of the wavelength.

Formula used:
$v = \sqrt {\dfrac{T}{\mu }}$
Here, T is the tension in the string and $\mu$ is the linear mass density of the string.

Complete step by step answer:
We know that when a string of length l is stretched with tension T, and a tuning fork or vibrator is used to make it vibrate, the transverse wave produces along the length of the string.
We have the velocity of this wave is,
$v = \sqrt {\dfrac{T}{\mu }}$
Here, T is the tension in the string and $\mu$ is the linear mass density of the string.
Therefore, we can write the above equation as follows,
$v = \sqrt {\dfrac{T}{{\left( {\dfrac{m}{l}} \right)}}}$
$\Rightarrow v = \sqrt {\dfrac{{Tl}}{m}}$
We can substitute 150 N for T, 150 cm for l and 400 mg for m in the above equation.
$v = \sqrt {\dfrac{{\left( {150\,N} \right)\left( {1.5\,m} \right)}}{{0.4 \times {{10}^{ - 3}}\,kg}}}$
$\therefore v = 750\,m/s$
We have to calculate the wavelength of the wave produced on the string to determine the frequency of the wave. We have given that the vibrators cause 6 loops in the string. Therefore, the length of the string is made up of 6 loops of length $\dfrac{\lambda }{2}$.
$l = 6\left( {\dfrac{\lambda }{2}} \right)$
$\Rightarrow \lambda = \dfrac{l}{3}$
Therefore, we get the wavelength of the wave,
$\lambda = \dfrac{{150\,cm}}{3}$
$\therefore \lambda = 50\,cm$
We know the relation between velocity, wavelength and frequency of the wave,
$f = \dfrac{v}{\lambda }$
We can substitute 750 m/s for v and 50 cm for $\lambda$ in the above equation.
$f = \dfrac{{750\,m/s}}{{50 \times {{10}^{ - 2}}\,m}}$
$\therefore f = 1.5\,kHz$
We know that, if $\lambda$ is the wavelength of the wave, then $\dfrac{\lambda }{2}$ is the distance between the two nodes or two antinodes.

Therefore, we can say that the distance between the two nodes is half of the wavelength of the wave that is 25 cm.

So, the correct answer is option (B) and (C).

Note:
Students should remember that the distance between two antinodes or two nodes is half of the wavelength of the wave. While using the formula $v = \sqrt {\dfrac{T}{\mu }}$, $\mu$ is the mass per unit length of the string and not just mass of the string.