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 the figure. The lowest frequency with which the heavy string resonates is $120\text{Hz}$. If the movable support is pushed to the right by $10\text{cm}$ so that the joint is placed on the pulley then the minimum frequency at which the heavy string can resonate will be

• A. $240\text{Hz}$
• B. $220\text{Hz}$
• C. $260\text{Hz}$
• D. $280\text{Hz}$

Answer 1

The correct option is A $240\text{Hz}$

Before the movable support is pushed, the heavy string will act as string with one end free.

So, the fundamental frequency of string is
$f=\frac{v}{4l}$

where, $v=$ speed of wave and $l=$ length of string.

From the data given in the question,

$120=\frac{v}{4l}\Rightarrow \frac{v}{l}=120\times 4=480..........\left(1\right)$

When the movable support is pushed by $10\text{cm}$ to the right, the joint is placed on the pulley and the heavy string will act as string with both ends fixed.

So, modes of vibration is given by $f=\frac{nv}{2l}$

For fundamentel frequency , $n=1$.

$\Rightarrow f=\frac{v}{2l}$ [for minimum frequency]

$\Rightarrow f=\frac{480}{2}=240\text{Hz}$ [from Eq. (1)]

Hence, option (a) is the correct answer.