A 2.00 kg mass is connected to a spring with a spring constant of 6.00 N/m. The displacement is given by the expression x(t) = (12.0 cm) sin(ω t). What is the maximum velocity of the mass?
Question
A 2.00 kg mass is connected to a spring with a spring constant of 6.00 N/m. The displacement is given by the expression x(t) = (12.0 cm) sin(ω t). What is the maximum velocity of the mass?
Solution
The maximum velocity of a mass-spring system in simple harmonic motion is given by the equation:
V_max = ω * A
where:
- ω is the angular frequency
- A is the amplitude of the motion
From the given displacement equation x(t) = (12.0 cm) sin(ω t), we can see that the amplitude A is 12.0 cm or 0.12 m.
The angular frequency ω can be found from the equation for the period of a mass-spring system:
T = 2π √(m/k)
where:
- T is the period
- m is the mass
- k is the spring constant
Rearranging for ω gives:
ω = √(k/m)
Substituting the given values:
ω = √(6.00 N/m / 2.00 kg) = √(3.00 s⁻²) = 1.73 rad/s
Finally, substituting these values into the equation for V_max gives:
V_max = 1.73 rad/s * 0.12 m = 0.208 m/s
So, the maximum velocity of the mass is 0.208 m/s.
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