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

The amplitude of the oscillator is 1.989 m.

Step-by-step explanation:

The amplitude of an oscillator can be determined using the initial displacement and initial velocity of the system. In this case, the initial displacement is given as 1.00 m and the initial velocity as 1.72 m/s. The amplitude, also known as the maximum displacement, is equal to the square root of the sum of the squares of the initial displacement and initial velocity.

Using the formula:

X = √(x₀² + v₀²)

Where X is the amplitude, x₀ is the initial displacement, and v₀ is the initial velocity.

Substituting the given values into the formula:

X = √(1.00² + 1.72²)

= √(1 + 2.9584)

= √3.9584

= 1.989 m

The amplitude of the oscillator is 1.989 m.

Answer 2

A=1.0.34 m

Step-by-step 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√(A^2-x^2)`

A=Amplitude

`1.72=2* \pi* √(A^2-1^2)`

`A^2-1=\left((1.72)/(2\pi)\right)^2`

`A^2=\left((1.72)/(2\pi)\right)^2+1`

`A^2=1.07\\A=√(1.07)\ m\\A=1.034\ m`

A=1.0.34 m