A heavy string is fastened at one end to a movable support and at...

Initially, the heavy string is fixed at the movable support and attached to a light thread at the other end. Since the light thread has negligible mass, it cannot support a transverse force. Therefore, the joint acts as a free end (an antinode) for the transverse waves in the heavy string.

For a string of length $L$ fixed at one end and free at the other, the lowest resonant (fundamental) frequency is given by:
$f_1 = \dfrac{v}{4L} = 120 \text{ Hz}$

When the movable support is pushed $10 \text{ cm}$ to the right, the entire heavy string shifts by $10 \text{ cm}$, causing the joint to rest exactly on the fixed pulley. The pulley restricts any transverse motion, so this end now acts as a fixed end (a node).

The heavy string is now fixed at both ends. Its length $L$ remains unchanged, and the tension $T$ (provided by the suspended weight) and mass per unit length $\mu$ also remain constant. Thus, the wave speed $v = \sqrt{\dfrac{T}{\mu}}$ is unchanged.

For a string fixed at both ends, the new lowest resonant frequency is:
$f_1' = \dfrac{v}{2L}$

Comparing the two frequencies, we get:
$f_1' = 2 \times \left(\dfrac{v}{4L}\right) = 2 \times f_1$

Substituting the given value:
$f_1' = 2 \times 120 \text{ Hz} = 240 \text{ Hz}$

Answer: $240 \text{ Hz}$