Question

A particle moving with SHM has velocities 4 cm/s and 3 cm/s at distances 3 cm and 4 cm respectively from equilibrium position. Find(a) the amplitude of oscillation (b) angular frequency (c) velocity of the particle as it passes through the equilibrium position.

Answer 1

Answer:

• The amplitude of oscillation is 5cm.

• Angular frequency is 1 rad/sec.

Given:

• V1 = 4 m/s at x = 3m
• V2 = 3 m/s at x = 4m

To Find:

• The amplitude of oscillation.
• Angular frequency.

Solution:

To find the amplitude = ?

We know that,

$\implies \sf v = \omega \sqrt{A^2 - x^2}$

• $\sf 4 = \omega \sqrt{A^2 - 3^2}$ ––(i)
• $\sf 3 = \omega \sqrt{A^2 - 4^2}$ ––(ii)

Dividing equation (ii) by (i),

$\implies \sf \dfrac{4 = \omega \sqrt{A^2 - 3^2}}{3 = \omega \sqrt{A^2 - 4^2}}$

$\implies \sf \dfrac{4}{3} = \dfrac{\omega \sqrt{A^2 - 9}}{\omega {\sqrt{A^2 - 16}}}$

$\implies \sf \dfrac{16}{9} = \dfrac{A^2 - 9}{A^2 - 16}$

Cross multiplication,

$\implies \sf 16(A^2 - 16) = 9(A^2 - 9)$

$\implies \sf 16A^2 - 256 = 9A^2 - 81$

$\implies \sf 7A^2 = 256 - 81$

$\implies \sf A^2 = \dfrac{175}{7}$

$\implies \sf A^2 = 25$

$\implies \sf A = \sqrt{25}$

$\implies \sf A = 5cm$

$\therefore$ Amplitude of oscillation is 5cm.

To find the angular frequency = ?

$\implies \sf v = \omega \sqrt{A^2 - x^2}$

$\implies \sf 3 = \omega \sqrt{5^2 - 4^2}$

$\implies \sf 3 = \omega \sqrt{25 - 16}$

$\implies \sf \omega = \dfrac{3}{3}$

$\implies \sf \omega = 1$

$\therefore$ Angular frequency is 1 rad/sec.

Answer 2

Answer:

• The amplitude of oscillation is 5cm.

• Angular frequency is 1 rad/sec.

Given:

V1 = 4 m/s at x = 3m

V2 = 3 m/s at x = 4m

To Find:

The amplitude of oscillation.

Angular frequency.

Solution:

To find the amplitude = ?

We know that,

$\implies \sf v = \omega \sqrt{A^2 - x^2}$

$\sf 4 = \omega \sqrt{A^2 - 3^2}$ ––(i)

$\sf 3 = \omega \sqrt{A^2 - 4^2}$ ––(ii)

Dividing equation (ii) by (i),

$\implies \sf \dfrac{4 = \omega \sqrt{A^2 - 3^2}}{3 = \omega \sqrt{A^2 - 4^2}}$

$\implies \sf \dfrac{4}{3} = \dfrac{\omega \sqrt{A^2 - 9}}{\omega {\sqrt{A^2 - 16}}}$

$\implies \sf \dfrac{16}{9} = \dfrac{A^2 - 9}{A^2 - 16}$

Cross multiplication,

$\implies \sf 16(A^2 - 16) = 9(A^2 - 9)$

$\implies \sf 16A^2 - 256 = 9A^2 - 81$

$\implies \sf 7A^2 = 256 - 81$

$\implies \sf A^2 = \dfrac{175}{7}$

$\implies \sf A^2 = 25$

$\implies \sf A = \sqrt{25}$

$\implies \sf A = 5cm$

$\therefore$ Amplitude of oscillation is 5cm.

To find the angular frequency = ?

$\implies \sf v = \omega \sqrt{A^2 - x^2}$

$\implies \sf 3 = \omega \sqrt{5^2 - 4^2}$

$\implies \sf 3 = \omega \sqrt{25 - 16}$

$\implies \sf \omega = \dfrac{3}{3}$

$\implies \sf \omega = 1$

$\therefore$ Angular frequency is 1 rad/sec.