An air-track glider is attached to a spring on a frictionless horizontal surface. The air-track glider is pulled to the right and released from rest at t = 0 s. If the air-track glider reaches a maximum speed of 43.0 cm/s and oscillates with a period of 4.00 s, what is the air-track glider's acceleration at t = 0.500 s? -1.74 cm/s2 -77.5 cm/s2 -86.0 cm/s2 -47.8 cm/s2
Question
An air-track glider is attached to a spring on a frictionless horizontal surface. The air-track glider is pulled to the right and released from rest at t = 0 s. If the air-track glider reaches a maximum speed of 43.0 cm/s and oscillates with a period of 4.00 s, what is the air-track glider's acceleration at t = 0.500 s? -1.74 cm/s2 -77.5 cm/s2 -86.0 cm/s2 -47.8 cm/s2
Solution
The motion of the air-track glider is simple harmonic motion (SHM). In SHM, the acceleration of the object is given by the equation a = -ω^2 * x, where ω is the angular frequency and x is the displacement.
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First, we need to find the angular frequency (ω). The angular frequency is related to the period (T) by the equation ω = 2π/T. Given that T = 4.00 s, we can calculate ω as follows:
ω = 2π/T = 2π/4.00 s = 1.57 rad/s
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At t = 0.500 s, the glider is at its maximum displacement (x_max), because it was released from rest at this position. The maximum displacement is related to the maximum speed (v_max) by the equation x_max = v_max/ω. Given that v_max = 43.0 cm/s, we can calculate x_max as follows:
x_max = v_max/ω = 43.0 cm/s / 1.57 rad/s = 27.39 cm
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Now, we can find the acceleration (a) at t = 0.500 s using the equation a = -ω^2 * x. Substituting the values we found for ω and x_max, we get:
a = -ω^2 * x_max = -(1.57 rad/s)^2 * 27.39 cm = -67.8 cm/s^2
So, the closest answer to this value is -77.5 cm/s^2.
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