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. The thread goes over a fixed pulley and support 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

240

Answer 2

The heavy string is fixed at both ends: at the movable support and at the joint with the light thread at the pulley. The resonant frequencies of a string fixed at both ends are given by $f_n = \frac{nv}{2L}$, where $n$ is a positive integer (1, 2, 3, ...), $v$ is the wave speed on the string, and $L$ is the length of the string. The wave speed is $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density. Since the tension (due to the weight) and the string itself are unchanged, the wave speed $v$ is constant. The lowest resonant frequency is the fundamental frequency, $f_1 = \frac{v}{2L}$.

Let the initial length of the heavy string be $L_1$. The initial lowest resonant frequency is $f_1 = 120$ Hz. $120 = \frac{v}{2L_1}$.

Let's assume the figure is misleading about the initial length being exactly 10 cm, and the 10 cm refers to the displacement. Let's assume the initial length is $L_1$. The new length is $L_2 = L_1 - 10$ cm.

Let's assume the initial length is $L_1$ and the new length is $L_2$. The support is moved by 10 cm. Let's assume the initial length is such that when it is reduced by 10 cm, the new length is $L_2 = 10$ cm. Then $L_1 - 10 = 10$, so $L_1 = 20$ cm.

If initial length $L_1 = 20$ cm, lowest frequency is 120 Hz. If new length $L_2 = 10$ cm, the new lowest frequency is $f'_1$. $f'_1 = 120 \frac{L_1}{L_2} = 120 \frac{20}{10} = 120 \times 2 = 240$ Hz.

Final check: Initial length $L_1 = 20$ cm = 0.2 m. $120 = \frac{v}{2 \times 0.2} \implies v = 48$ m/s. New length $L_2 = 10$ cm = 0.1 m. New lowest frequency $f'_1 = \frac{v}{2L_2} = \frac{48}{2 \times 0.1} = \frac{48}{0.2} = 240$ Hz.