Question

The tension in the wire is decreased by 19 %. What will be the percentage decrease in frequency?
${\text{A}}{\text{.}}$ 0.19 %
${\text{B}}{\text{.}}$ 10 %
${\text{C}}{\text{.}}$ 19 %
${\text{D}}{\text{.}}$ 0.9 %

Answer 1

Hint: Here, we will proceed by writing down a relation between the frequency of oscillation and the tension in the wire. Then, we will express this equation in terms of percentage decrease of frequency and percentage decrease of tension in the wire.

Formulas Used- ${\text{f}} = \dfrac{1}{{2{\text{l}}}}\sqrt {\dfrac{{\text{T}}}{\mu }} $ and $y = k{\left( x \right)^n} \Rightarrow \dfrac{{\Delta y}}{y} = n\left( {\dfrac{{\Delta x}}{x}} \right)$.
Given, Percentage decrease in tension T = 19 %

Step By Step Answer:
As we know that the frequency of oscillation of any wire having tension in it is given by

${\text{f}} = \dfrac{1}{{2{\text{l}}}}\sqrt {\dfrac{{\text{T}}}{\mu }} {\text{ }} \to {\text{(1)}}$ where f denotes the frequency of oscillation, l denotes the length of the wire, T denotes the tension in the wire and $\mu $ denotes the mass per unit length of the wire (i.e., $\mu = \dfrac{{\text{m}}}{{\text{l}}}$)

As given in the problem, the tension in the wire T is changing and corresponding to which we have to find the change in frequency (which means frequency will also be changing)

Here, the length of the wire is not changed (i.e., length of the wire l is constant). Also, mass per unit length for the wire will also not change (i.e., $\mu $ will also be constant)

Equation (1), can be rewritten as under

${\text{f}} = \left( {\dfrac{1}{{2{\text{l}}}}\sqrt {\dfrac{{\text{1}}}{\mu }} } \right)\sqrt {\text{T}} $

In the above equation, $\left( {\dfrac{1}{{2{\text{l}}}}\sqrt {\dfrac{{\text{1}}}{\mu }} } \right)$ = k (any constant), we will get

$
  {\text{f}} = k\sqrt {\text{T}} \\
  {\text{f}} = k{\left( {\text{T}} \right)^{\dfrac{1}{2}}}{\text{ }} \to {\text{(2)}} \\
 $

As we know that any for any equation $y = k{\left( x \right)^n}{\text{ }} \to {\text{(3)}}$ in which y and x are variables and are changing whereas k is a constant, we can write

$\dfrac{{\Delta y}}{y} = n\left( {\dfrac{{\Delta x}}{x}} \right){\text{ }} \to {\text{(4)}}$

By comparing equations (2) and (3), we can say that the frequency of oscillation f and tension in the wire T are both variables where y = f, n = $\dfrac{1}{2}$ and x = T

By replacing y with f and x with T in the formula given in equation (4), we get

\[\dfrac{{\Delta {\text{f}}}}{{\text{f}}} = \dfrac{1}{2}\left( {\dfrac{{\Delta {\text{T}}}}{{\text{T}}}} \right){\text{ }} \to {\text{(5)}}\]

Also we know that the percentage decrease in any quantity x can be represented as

Percentage decrease in x (in %) = \[ - \left( {\dfrac{{\Delta x}}{x}} \right) \times 100{\text{ }} \to {\text{(6)}}\]

By multiplying -100 on both the sides of equation (5), we get

\[
   - \left( {\dfrac{{\Delta {\text{f}}}}{{\text{f}}}} \right) \times 100 = - \dfrac{1}{2}\left( {\dfrac{{\Delta {\text{T}}}}{{\text{T}}}} \right) \times 100 \\
   - \left( {\dfrac{{\Delta {\text{f}}}}{{\text{f}}}} \right) \times 100 = \dfrac{1}{2}\left[ { - \left( {\dfrac{{\Delta {\text{T}}}}{{\text{T}}}} \right) \times 100} \right] \\
 \]

Using the formula given by equation (6) in above equation, we get

Percentage decrease in f = $\dfrac{1}{2}$(Percentage decrease in T)

$ \Rightarrow $ Percentage decrease in f = $\dfrac{1}{2}$(19) % = 9.5 %
This percentage decrease in tension in the wire is approximately 10 %.

Hence, option B is correct.

Note: Percentage change in any quantity is simply the ratio of the change occurred in that quantity to the value of that quantity before the change (i.e., initial value of that quantity). The negative sign in \[ - \left( {\dfrac{{\Delta x}}{x}} \right) \times 100\] represents that this is a percentage decrease in quantity x. Percentage increase in any quantity x is represented as \[\left( {\dfrac{{\Delta x}}{x}} \right) \times 100\].