Question

A chord attached about an end to a vibrating fork divides it into $6$ loops, when its tension is $36\,{\text{N}}$ . The tension at which it will vibrate in $4$ loops is:
(A) $24\,{\text{N}}$
(B) $36\,{\text{N}}$
(C) $64\,{\text{N}}$
(D) $81\,{\text{N}}$

Answer 1

First of all, we will find a relation between the wavelength and length of the string for both the cases. Two loops form a wave. We will manipulate the two equations followed by substitution of required values to obtain the result.

Complete step by step answer:
In the given question, we are supplied with the following data:
There are six loops in the chord which is attached about an end to a vibrating fork.
The tension in the first case is given as $36\,{\text{N}}$ .
In the second case, there are four loops in the string.
We are asked to find the tension force in the string which was attached about the end of the vibrating fork.
To begin with, let us make it very clear that, when a string vibrates, there creates a wave which is composed of a crest and a trough. Crest is the upward displacement while trough is the downward displacement. We can say that a wave contains two loops and the length of a complete wave is called wavelength.
So, we can say that the wavelength of the wave is equivalent to two loops.
Mathematically we can write:
For the first case,
Since, $2$ loops are equivalent to $1{\lambda _1}$ .
So, $6$ loops are equivalent to $3{\lambda _1}$ .
Now, we can write:
L = 3{\lambda _1} \\\ \Rightarrow {\lambda _1} = \dfrac{L}{3} \\\
Where,
$L$ indicates the length of the string.
For the second case,
Since, $2$ loops are equivalent to $1{\lambda _2}$ .
So, $4$ loops are equivalent to $2{\lambda _2}$ .
Now, we can write:
L = 2{\lambda _2} \\\ \Rightarrow {\lambda _2} = \dfrac{L}{2} \\\
For all the waves along a string, we can write:
$v \propto \sqrt T$
$\Rightarrow \lambda \propto \sqrt T$ …… (1)
Where,
$\lambda$ indicates the wavelength.
$T$ indicates the tension force in the string.
For the first case the equation (1) becomes:
$\Rightarrow {\lambda _1} = \sqrt {36}$ …… (2)
For the second case the equation (1) becomes:
$\Rightarrow {\lambda _2} = \sqrt {{T_2}}$ …… (3)
Now, we divide the equations (2) and (3), and we get:
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\sqrt {36} }}{{\sqrt {{T_2}} }}$ …… (4)
We substitute the required values in the equation (4) and we get:
$\Rightarrow \dfrac{{\left( {\dfrac{L}{3}} \right)}}{{\left( {\dfrac{L}{2}} \right)}} = \dfrac{{\sqrt {36} }}{{\sqrt {{T_2}} }} \\\ \Rightarrow \dfrac{2}{3} = \dfrac{{\sqrt {36} }}{{\sqrt {{T_2}} }} \\\ \Rightarrow {\left( {\dfrac{2}{3}} \right)^2} = \dfrac{{36}}{{{T_2}}} \\\ \Rightarrow {T_2} = \dfrac{{36 \times 9}}{4} \\\ \Rightarrow {T_2} = \dfrac{{36 \times 9}}{4} \\\ \Rightarrow {T_2} = 9 \times 9 \\\ \therefore {T_2} = 81\,{\text{N}}$
Hence, the tension force in the string which was attached about the end of the vibrating fork is $81\,{\text{N}}$ .

The correct option is D.

Note: While solving the problem, it is important to remember that the loops that have been mentioned here in the question are nothing but the crests and the troughs. In a complete wave, there are two loops in it. Again, most of the students seem to have some sort of confusion regarding the length of the string. As in the question there is no mention of anything about the change in length of the string, so in those cases we take the length of the string equal in both the cases.