Question

A heavy string is tied at one end to a movable support and to a light thread at the other end as shown in figure (15-E12). The thread goes over a fixed pulley and supports a weight to produce a tension. The lowest frequency with which the heavy string resonates is 120 Hz. If the movable support is pushed to the right by 10 cm so that the joint is placed on the pulley, what will be the minimum frequency at which the heavy string can resonate?

Answer 1

Let T be the tension in the string and m be the mass per unit length of the heavy string.
In the first part of the question, the heavy string is fixed at only one end.
So, the lowest frequency is given by:
$f_0=\frac{1}{4L}\sqrt{\frac{T}{m}}...\left(1\right)$
When the movable support is pushed by 10 cm to the right, the joint is placed on the pulley and the heavy string becomes fixed at both the ends (keeping T and m same).
Now, the lowest frequency is given by:
$f_0\text{'}=\frac{1}{2L}\sqrt{\frac{T}{m}}...\left(2\right)$
Dividing equation (2) by equation (1), we get:
$f_0\text{'}=2f_0=240\mathrm{Hz}$