Question

An oscillator with angular frequency of 1.00 s-1has initial displacement of 1.00 m and initial velocity of 1.72 m/s. What is the amplitude of oscillation?

Answer 1

Answer:

A=1.0.34 m

Explanation:

Given that

f= 1 s

x= 1 m

v = 1.72 m/s

We know that angular frequency ω is given as

ω = 2 π f

Now by putting the values in the above equation

ω = 2 π x 1

ω=  2 π  rad/s

The velocity v is given as

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

A=Amplitude

$1.72=2\times \pi\times \sqrt{A^2-1^2}$

$A^2-1=\left(\dfrac{1.72}{2\pi}\right)^2$

$A^2=\left(\dfrac{1.72}{2\pi}\right)^2+1$

$A^2=1.07\\A=\sqrt{1.07}\ m\\A=1.034\ m$

A=1.0.34 m