A 1.00 kg glider attached to a spring with a spring constant of 40 N/m oscillates on a horizontal, frictionless air track. At t = 0 the glider is released from rest at x = -17 cm. Find the maximum value of the velocity in m/s.
To find the maximum value of the velocity of the glider, we need to determine the maximum displacement (amplitude) of the oscillation, and then use the relationship between displacement and velocity in simple harmonic motion.
First, let's determine the amplitude of the oscillation. We are given that the glider is released from rest at x = -17 cm (or -0.17 m). In simple harmonic motion, the amplitude is simply the maximum displacement from the equilibrium position. Since the glider is released from rest, its initial position is the equilibrium position, so the amplitude is 0.17 m.
Next, we can use the formula for simple harmonic motion to find the maximum velocity. In simple harmonic motion, the velocity is given by the equation:
v = ω * A
where v is the velocity, ω is the angular frequency, and A is the amplitude.
The angular frequency, ω, can be calculated using the formula:
ω = sqrt(k / m)
where k is the spring constant and m is the mass of the glider.
Given that the spring constant, k, is 40 N/m and the mass, m, is 1.00 kg, we can calculate the angular frequency:
ω = sqrt(40 N/m / 1.00 kg)
= sqrt(40) rad/s
≈ 6.32 rad/s
Finally, we can substitute the values of ω and A into the velocity formula to find the maximum velocity:
v = ω * A
≈ 6.32 rad/s * 0.17 m
≈ 1.07 m/s
Therefore, the maximum value of the velocity of the glider is approximately 1.07 m/s.