Question

The tension in a wire is decreased by $19\%$. The percentage decrease in frequency will be
(a) $10\%$
(b) $19\%$
(c) $0.19\%$
(d) None of the these

Answer 1

Hint: The relation between tension in the string and frequency is given by the formula, $f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu }}$. Now, by assuming the original tension as T we will find the value of new tension after decrease and then we will substitute them in expression of frequency. Then, by taking the ratios of original and final frequencies we will find the value of final frequency. Then by using the formula of percentage change i.e. $\text{percentage change}=\dfrac{final-original}{original}\times 100\%$, we will find the percentage change in frequency.

Formula used: $f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu }}$, $\text{percentage change}=\dfrac{final-original}{original}\times 100\%$

Complete step by step answer:
Now, in the question it is given that the tension in a wire is decreased by $19\%$, so, to find the decrease in frequency we will use the formula,
$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu }}$
Where, f is frequency, T is tension, l is length of string and \[\mu \] is mass per unit length.
Now, we will consider that the length and mass per unit length of the spring will be constant so we will neglect those terms, then the equation will be,
$f\alpha \sqrt{T}$
$f\alpha {{\left( T \right)}^{\dfrac{1}{2}}}$ ………….(i)
Now, considering the change in frequency as $f'$ and change in tension as $T'$, we will find the change in frequency. Now, the tension decreases by $19\%$, so we can say that fi original tension was 100 then change in the new tension can be given as,
$T'=\dfrac{\left( 100-19 \right)}{100}T=\dfrac{81}{100}T$
Now, the ne tension based on new frequency can be given as,
$f'={{\left( T' \right)}^{\dfrac{1}{2}}}$ ……………(ii)
Taking ratios of expression (i) and (ii) we will get,
$\dfrac{f'}{f}={{\left( \dfrac{T'}{T} \right)}^{\dfrac{1}{2}}}$
Now, keeping the value of $T'$ in expression we will get,
$\dfrac{f'}{f}={{\left( \dfrac{\dfrac{81}{100}T}{T} \right)}^{\dfrac{1}{2}}}$
$\dfrac{f'}{f}={{\left( \dfrac{81}{100} \right)}^{\dfrac{1}{2}}}$
$f'=\dfrac{9}{10}f$
Now, percentage change can be given by formula,
$\text{percentage change}=\dfrac{final-original}{original}\times 100\%$
Now, using this formula, the change in frequency can be given as,
$\text{percentage change}=\dfrac{f'-f}{f}\times 100\%$
Now, on substituting the value of $f'$, we will get,
$\text{percentage change}=\dfrac{\dfrac{9}{10}f-f}{f}\times 100\%=\dfrac{\left( 9-10 \right)f}{10f}\times 100\%$
$\text{percentage change=}\dfrac{-1}{10}\times 100\%=-10\%$
Here, negative sign indicates the decrease in frequency i.e. $10\%$.

So, the correct answer is “Option A”.

Note: Students might consider the formula of percentage change as $\text{percentage change}=\dfrac{original-final}{original}\times 100\%$ instead of $\text{percentage change}=\dfrac{final-original}{original}\times 100\%$ and due to that on substituting the values we will get, $\text{percentage change}=\dfrac{f-\dfrac{9}{10}f}{f}\times 100\%=\dfrac{1}{10}\times 100\%=10\%$, then also it is not wrong because by both the ways answer remains same. So, it can be considered as an alternate method to find the answer.