Question

The length of a sonometer wire AB is 110 cm. The distance at which two bridges should be placed from A to divide the wire into 3 segments whose fundamental frequencies are in ratio 1:2:3?
A. 30 cm
B. 60 cm, 30 cm, 20 cm
C. 80 cm
D. 40 cm, 80 cm

Answer 1

Hint: A sonometer consists of a string attached by some weight (providing tension to the string) on top of a wooden box. When the tension and mass of the string remains unchanged, the frequency n of the fundamental mode is inversely proportional to the length of the string.

Formula used:
The frequency of fundamental mode is inversely proportional to length:
$ n \propto \dfrac{1}{l}$

Complete answer:
A sonometer has adjustable two wedges attached to it (called as bridges in the question). We are given a 110 cm long string. If we place the bridges such that the string gets divided into three segments of lengths $l_1$, $l_2$ and $l_3$ then we will get frequencies with fundamental modes $n_1$, $n_2$ and $n_3$ respectively from these segments.
In the question, we are already given that frequencies of these three segments of string are in ratio 1:2:3 so we may write:
$ n_1 : n_2 : n_3 = 1:2:3$
Now, we know that n is inversely proportional to the length of the segments so we may write:
$\dfrac{1}{l_1} : \dfrac{1}{l_2} : \dfrac{1}{l_3} = 1:2:3$
If we just take reciprocal on both sides, we get:
$l_1 : l_2 : l_3 = \dfrac{1}{1} : \dfrac{1}{2} : \dfrac{1}{3}$
Taking LCM of the RHS (which is 6) and multiplying the LCM to it, we get:
$l_1 : l_2 : l_3 = 6 : 3 : 2 $

So, we obtained the ratio for the lengths. We are also given the total length of the string. So, we just use the proportions here and write:
$l_1 = \left( \dfrac{6}{6+3+2} \right) . 110$ cm = 60 cm
$l_2 = \left( \dfrac{3}{6+3+2} \right) . 110$ cm = 30 cm
$l_3 = \left( \dfrac{2}{6+3+2} \right) . 110$ cm = 20 cm
Thus we get the lengths of the three segments.
Now, we are asked that at what distance from A are the bridges to be placed.
Since, $l_1$= 60 cm, the first bridge should be placed at this length from the point A.
Now, $l_2$= 30 cm, so the second bridge should be placed at a length of (60 cm+30 cm =) 90cm from A.

So, the correct answer is “Option B”.

Note:
The frequency of the fundamental mode of a sonometer is directly proportional to the square root of the tension on the string and inversely proportional to the square root of mass per unit length of the string. Since we held these two quantities as constant in our case the frequency only depended on the length of the string. To remember this better one can also think about a guitar. In a guitar, as we tune it by reducing the length of its string, different frequencies are heard.