Question

One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates at 120 Hz. The other end passes over a pulley and supports a 1.50 kg mass. The linear mass density of the rope is 0.0750 kg/m. The speed of a transverse wave on the rope is $1.41\times {{10}^{x}}m/s$ . Find the value of x.

Answer 1

Hint: For a rope, the speed is given by: $v=\sqrt{\dfrac{T}{\mu }}$ , where T is the tension force and $\mu $ is the linear mass density. Here, for a given rope, the tension force is equal to $T=mg$, where m is mass and g is the acceleration dur to gravity. Using this formula, find the tension force and then substitute the value of T in the formula $v=\sqrt{\dfrac{T}{\mu }}$to find the transverse velocity.

Complete step by step answer:We have:
$\begin{align}
  & f=120Hz \\
 & m=1.5kh \\
 & \mu =0.075kg/m \\
\end{align}$
As we know that: $T=mg$
We get:
$T=1.5\times 10N......(1)$
Now, we know that $v=\sqrt{\dfrac{T}{\mu }}.......(2)$
So, substitute the value of tension force from equation (1) in equation (2), we get:
$\begin{align}
  & v=\sqrt{\dfrac{1.5\times 10}{0.075}} \\
 & =14.1m/s \\
 & =1.41\times {{10}^{1}}m/s......(3)
\end{align}$
Since, velocity is in terms of $1.41\times {{10}^{x}}m/s$
So, by comparing this value with equation (3), we get x = 1.
Hence, the value of x = 1 in the transverse velocity.

Note:In the formula: $v=\sqrt{\dfrac{T}{\mu }}$, T represents tension force in the rope. Do not confuse it with the time period. Since, we are given the value of frequency in the question. So, be careful with the formula, if the value is required to solve or it is just for confusion sake.