SATHEE: Simple Harmonic Motion 3 Question 15

Simple Harmonic Motion 3 Question 15

15. A $0.1 k g$ 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 $140 r a d s^{- 1}$ . If the Young’s modulus of the material of the wire is $n \times 10^{9} N m^{- 2}$ , the value of $n$ is.

(2010)

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Answer:

Correct Answer: 15. 4

Solution:

$ω = \sqrt{\frac{K}{m}} = \sqrt{\frac{Y A}{l m}} = \sqrt{\frac{(n \times 10^{9}) (4.9 \times 10^{- 7})}{1 \times 0.1}}$

Putting, $ω = 140 r a d s^{- 1}$ in above equation we get, $n = 4$

$∴$ Answer is 4 .

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Simple Harmonic Motion 3 Question 15

Answer:

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