Question

let us consider a system of a spherical bob attached to the bottom end of a vertical spring. The oscillations set into in this system with an amplitude of 0.2m and the value spring constant is 16N/m then what be the weight of bob if the maximum speed of bob 0.4m/s?​

Answer 1

The weight of the bob is 39.2 N

Explanation:

Given:

A vertical spring has amplitude 0.2 m, the value of spring constant k is 16 N/m

The maximum speed of the bob is 0.4 m/s

To find out:

The weight of the bob

Solution:

Let the mass of the bob is m

Given that

A = 0.2 m

k = 16 N/m

$v_{max}=0.4$ m/s

The angular frequency in a simple harmonic motion is given by

$\omega=\sqrt{k/m}$

$\implies \omega=\sqrt{\frac{16}{m}}$

$\implies \omega=\frac{4}{\sqrt{m}}$

Also, in a simple harmonic motion

$v_{max}=\omega A$

Thus,

$v_{max}=\frac{4}{\sqrt{m}}A$

$\implies 0.4=\frac{4}{\sqrt{m}}\times 0.2$

$\implies 0.1\sqrt{m}=0.2$

$\implies \sqrt{m}=2$

$\implies m=4$ kg

Thus, weight of the bob = mg = 4 × 9.8 = 39.2 N

Hope this answer is helpful.

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