Q99 / 160 · JIPMER 2014 · Physics

Linear density of a string of is $1.3 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ and wave equation is $y=0.021 \sin (x+30 t)$. Find, the tension in the string where, $x$ in metre and $t$ in second.

  • Correct answer
    A
    $0.12$ N
  • B
    $0.21$ N
  • C
    $1.2$ N
  • D
    $0.012$ N

Solution

Mass per unit length $m=1.3 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ $y=0.021 \sin (x+30 t)$ Comparing with $y=a \sin \left(\omega t+\frac{2 \pi x}{\lambda}\right)$ $a=0.021, \omega=30$ $2 \pi n=30$ $\begin{aligned} & \Rightarrow \quad n=\frac{30}{2 \pi} \\ & \text { We know that } \frac{2 \pi}{\lambda}=1 \\ & \end{aligned}$ $\begin{aligned} \therefore \quad \lambda & =\frac{1}{2 \pi} \\ \text { Frequency, } \quad n & =\frac{1}{\lambda} \sqrt{\frac{T}{m}} \\ \frac{30}{2 \pi} & =\frac{1}{2 \pi} \sqrt{\frac{T}{1.3 \times 10^{-4}}} \\ T & =1.3 \times 10^{-4} \times 900 \\ & =0.117=0.12 \mathrm{~N}\end{aligned}$

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

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