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 45.0 cm/s and oscillates with a period of 4.00 s, what is the object's position at t = 0.950 s?
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 45.0 cm/s and oscillates with a period of 4.00 s, what is the object's position at t = 0.950 s?
Solution
The motion of an object attached to a spring is simple harmonic motion, which can be described by the equation:
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 (maximum displacement from the equilibrium position),
- ω is the angular frequency of the motion,
- φ is the phase constant, which depends on the initial conditions.
Given that the object is released from rest at its maximum displacement, we know that φ = 0.
The maximum speed of the object is given by:
v_max = ωA
We can rearrange this to find the amplitude:
A = v_max / ω
The angular frequency ω is related to the period T of the motion by:
ω = 2π / T
Substituting the given values:
ω = 2π / 4.00 s = 1.57 rad/s
A = (45.0 cm/s) / (1.57 rad/s) = 28.7 cm
Now we can find the position of the object at t = 0.950 s by substituting these values into the equation for x(t):
x(0.950 s) = (28.7 cm) cos[(1.57 rad/s)(0.950 s)] = 28.7 cm * cos(1.49 rad) = 9.6 cm
So, the object's position at t = 0.950 s is 9.6 cm.
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