Q179 / 180 · COMEDK 2018 · Physics

A string vibrates with a frequency of $200 \mathrm{~Hz}$. When its length is doubled and tension is altered, it begins to vibrate with a frequency of $300 \mathrm{~Hz}$. The ratio of the new tension to the original tension is

  • Correct answer
    A
    9:1
  • B
    $1: 9$
  • C
    $3: 1$
  • D
    $1: 3$

Solution

Given, $L_{2}=2 L_{1}, f_{1}=200 \mathrm{~Hz}, f_{2}=300 \mathrm{~Hz}$ We know that, frequency of vibrating string is given as $$ f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} $$ where, $\mu=$ mass per unit length of string and $L=$ length of the string. $$ \begin{array}{ll} \Rightarrow & f \propto \frac{\sqrt{T}}{L} \\ \Rightarrow \quad & \frac{f_{2}}{f_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}, \frac{L_{1}}{L_{2}} \\ \Rightarrow \quad \sqrt{\frac{T_{2}}{T_{1}}}=\frac{f_{2} L_{2}}{f_{1} L_{1}}=\frac{300}{200} \times \frac{2 L_{1}}{L_{1}} \\ \Rightarrow \quad & \sqrt{\frac{T_{2}}{T_{1}}}=3 \\ & \frac{T_{2}}{T_{1}}=9 \\ \therefore \quad & T_{2}: T_{1}=9: 1 \end{array} $$

Hence, the correct answer is option: 'A'.

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