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A mass is oscillating on a spring with a period of 4.60 s. At t = 0 s the mass has zero speed and is at x = 8.30 cm. What is its speed at t = 2.50 s?

the answer is 3.06 cm/s but why/how?

User Magne Land
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1 Answer

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Step-by-step explanation:

To find the speed of the mass at t = 2.50 s, we can use the equation for simple harmonic motion:

x(t) = A * cos(ωt + φ)

where:

x(t) = displacement of the mass from equilibrium at time t

A = amplitude of the oscillation

ω = angular frequency of the oscillation

φ = phase angle (initial phase)

Given that the period T = 4.60 s, we know that the angular frequency is related to the period as follows:

ω = 2π / T

Substitute the given period T = 4.60 s to find ω:

ω = 2π / 4.60 s ≈ 1.363 s^(-1)

At t = 0 s, the mass has zero speed and is at x = 8.30 cm. This means that the initial phase angle φ = 0.

Now, we can find the displacement x at t = 2.50 s:

x(2.50 s) = A * cos(ω * 2.50 s)

We know that at t = 0 s, x = 8.30 cm, so:

8.30 cm = A * cos(0)

Since the cosine of 0 is 1, we can find the amplitude A:

A = 8.30 cm

Now, calculate the displacement at t = 2.50 s:

x(2.50 s) = 8.30 cm * cos(1.363 s^(-1) * 2.50 s) ≈ 8.30 cm * cos(3.4075) ≈ 3.06 cm

The displacement at t = 2.50 s is approximately 3.06 cm. However, the speed is the magnitude of the derivative of the displacement with respect to time:

v(t) = d(x(t))/dt = -A * ω * sin(ωt)

Substitute the known values:

v(2.50 s) = -8.30 cm * 1.363 s^(-1) * sin(1.363 s^(-1) * 2.50 s) ≈ -8.30 cm * 1.363 s^(-1) * sin(3.4075) ≈ -8.30 cm * 1.363 s^(-1) * (-0.2357) ≈ 3.06 cm/s

So, the speed of the mass at t = 2.50 s is approximately 3.06 cm/s.

User Lmcarreiro
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