An object is attached to a spring on a frictionless horizontal surface. The object is pulled to the right and released from rest at t = 0 s. If the object reaches a maximum speed of 43.5 cm/s and oscillates with a period of 5.00 s, what is the object's position at t = 0.600 s? 25.2 cm 34.6 cm 28.6 cm 23.7 cm
Question
An object is attached to a spring on a frictionless horizontal surface. The object is pulled to the right and released from rest at t = 0 s. If the object reaches a maximum speed of 43.5 cm/s and oscillates with a period of 5.00 s, what is the object's position at t = 0.600 s? 25.2 cm 34.6 cm 28.6 cm 23.7 cm
Solution
The motion of an object attached to a spring is simple harmonic motion. The equation for simple harmonic motion is:
x(t) = A cos(ωt + φ)
where:
- x(t) is the position of the object at time t,
- A is the amplitude of the motion,
- ω is the angular frequency of the motion,
- t is the time,
- φ is the phase constant.
Given that the object is released from rest at t = 0 s, the phase constant φ is 0.
The maximum speed of the object is given when the spring is at the equilibrium position. This occurs when the potential energy is minimum and the kinetic energy is maximum. The maximum speed v_max is related to the amplitude A and the angular frequency ω by the equation:
v_max = Aω
We can rearrange this equation to solve for the amplitude A:
A = v_max / ω
The period T of the motion is related to the angular frequency ω by the equation:
ω = 2π / T
We can substitute the given values into these equations to find the amplitude A and the angular frequency ω:
A = (43.5 cm/s) / (2π / 5.00 s) = 34.6 cm
ω = 2π / 5.00 s = 1.26 s^-1
Finally, we can substitute these values into the equation for simple harmonic motion to find the position of the object at t = 0.600 s:
x(0.600 s) = (34.6 cm) cos[(1.26 s^-1)(0.600 s) + 0] = 28.6 cm
So, the object's position at t = 0.600 s is 28.6 cm.
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