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A 200 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz . At one instant, the mass is at x=5.0cm and has vx=−30cm/s.Express your answer using two significant figures.

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User Effkay
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To solve this problem, we can use the equation for the frequency of an oscillating mass-spring system:

f = 1 / T

where f is the frequency and T is the period.

Given that the frequency is 2.0 Hz, we can find the period:

T = 1 / f
T = 1 / 2.0 Hz
T = 0.5 s

The velocity of the mass, vx, at any point can be determined using the equation:

vx = -Aωsin(ωt + φ)

where A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.

In this case, the mass is at x = 5.0 cm, so we can find the amplitude:

A = 5.0 cm = 0.05 m

The angular frequency can be calculated using the formula:

ω = 2πf

ω = 2π * 2.0 Hz
ω ≈ 12.57 rad/s

Now we can use the given velocity to find the phase constant:

vx = -Aωsin(ωt + φ)
-30 cm/s = -0.05 m * 12.57 rad/s * sin(12.57 rad/s * t + φ)

Since sin(12.57 rad/s * t + φ) = -1, we have:

-30 cm/s = -0.05 m * 12.57 rad/s * (-1)
-30 cm/s = 0.6285 m^2/s * m/s
-30 cm/s ≈ -0.63 m^2/s

To express the answer using two significant figures, we round the value to -0.63 m^2/s.

Therefore, the velocity of the mass is approximately -0.63 m^2/s.
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User Spinfire
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