Question

A 0.25 kg harmonic oscillator has a total mechanical energy of 4.1 j. if the oscillation amplitude is 20.0 cm, what is the oscillation frequency?

Answer 1

When the oscillator is at maximum extension, we know all of its energy is in Potential Energy, so if the total oscillation energy is 4.1 J, we know that at maximum displacement of 0.2 m, that 

energy = 1/2 kA^2 where A= 0.2 m 

k= 2E  / A^2 = 2*4.1 J /0.2^2=200 N/m 

the frequency of oscillation is (1/2pi) sqrt[k/m] 

knowing k and m, we can substitute values and find frequency

Answer 2

Answer:

4.6Hz

Explanation:

We first have to derive Oscillation Frequency formula

Maximum velocity = Vmax = wA

Energy = 1/2m(Vmax)²

Energy = 1/2m(wA)²

Frequency = w/2π

Hence,

Oscillation Frequency Formula is given as

F =( 1/2πA)× √2E/m

where in the question,

A = Amplitude = 20.0cm

We convert 20.0cm to meter

100cm = 1m

20cm = ?

= 20÷100 = 0.20m

E = Mechanical energy = 4.1j

m = mass = 0.25kg

Frequency =( 1/2π × 0.20m) × √(2×4.1j)/0.25kg

Frequency = 4.558Hz

Oscillatory frequency approximately is = 4.6Hz