Question

A mass of 0.5kg causes an extension 0.03mm in a spring and the system is set for oscillations
find - 1. Force constant
2. angular frequency ​

Answer 1

Answer:

I have no idea this question

Answer 2

Given:

A mass of 0.5kg causes an extension 0.03mm in a spring and the system is set for oscillations.

To find:

• Force constant
• Angular Frequency

Calculation:

Let force constant be "k":

$\therefore \: F = - kx$

$\implies \: mg= - kx$

$\implies \: 0.5 \times 10= - k( - 0.03 \times {10}^{ - 3} )$

$\implies \: 5= - k( - 0.03 \times {10}^{ - 3} )$

$\implies \: 5= k( 3\times {10}^{ - 5} )$

$\implies \:k = \dfrac{5}{3} \times {10}^{5} \: N/m$

$\boxed{ \implies \:k = 1.6 \times {10}^{5} \: N/m}$

Let angular frequency be $\omega$.

$\therefore \: \omega = \sqrt{ \dfrac{k}{m} }$

$\implies \: \omega = \sqrt{ \dfrac{1.6 \times {10}^{5} }{0.5} }$

$\implies \: \omega = \sqrt{3.2 \times {10}^{5} }$

$\implies \: \omega = \sqrt{32 \times {10}^{4} }$

$\implies \: \omega = 5.6 \times {10}^{2}$

$\implies \: \omega = 560 \: hz$

So, angular frequency is 560 Hz.