Question

A 2. 00 kg, frictionless block is attached to an ideal spring with force constant 300 n/m. At t = 0 the block has velocity -4. 00 m/s and displacement 0. 200 m away from equilibrium

Answer 1

The amplitude is 0.383 m and the phase angle is 1.02 rad.

What is amplitude?

The amplitude is the maximum displacement from the mean position.

Given is the mass m= 2kg and force constant k = 300 N/m, then the angular frequency will be

ω = √(k/m)

Putting the values, we get

ω =√(300/2) = √150 rad/s.

The general form of SHM equation is y = Acos(ωt +Ф)

differentiating w.r.t to time , we get the velocity, v = -Aωsin(ωt +Ф)

At t=0, y₀ = AcosФ and v₀ = -AωsinФ

Given, y₀ =0.2 m and v₀ = -4m/s

then,  AcosФ =0.2...............(1)

and

AωsinФ=4 or AsinФ=4/√150...............(2)

Squaring and adding both the equation, we get the Amplitude.

A = 0.383 m

Dividing both the equations, we get the phase angle,

tan Ф = 4/(0.2 x√150) =20/√150

Ф=1.02 rad

Thus, the amplitude is 0.383 m and the phase angle is 0.383m and  1.02 rad.

Learn more about Amplitude.

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