Simple Harmonic Motion Ques 23
- A $0.1 kg$ mass is suspended from a wire of negligible mass. The length of the wire is $1 m$ and its cross-sectional area is $4.9 \times 10^{-7} m^{2}$. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140 rad s^{-1}$. If the Young’s modulus of the material of the wire is $n \times 10^{9} Nm^{-2}$, the value of $n$ is.
(2010)
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Answer:
Correct Answer: 23.4
Solution:
Formula:
- $\omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{Y A}{l m}}=\sqrt{\frac{\left(n \times 10^{9}\right)\left(4.9 \times 10^{-7}\right)}{1 \times 0.1}}$
Putting $\omega=140\ \text{rad}\ \text{s}^{-1}$ in the above equation we get, $n=4$
$\therefore$ Answer is 4 .
BETA
