Question

An air-track glider oscillates horizontally on a spring at a frequency of 0.50 Hz. Suppose the glider is pulled to the right of its equilibrium position by 12 cm and then released. Where will the glider be 1.0 s after its release? What is its velocity at this point?

Answer 1

The glider will be 0.12 m to the left of its equilibrium position and have a velocity of -0.75 m/s after 1.0 second.

Step-by-step explanation:

The glider's position after 1.0 second can be determined using the equation for simple harmonic motion:

x = A * cos(2πft),

where x is the displacement from equilibrium, A is the amplitude, f is the frequency, and t is the time. In this case, the glider is pulled to the right by 12 cm, which corresponds to an amplitude of 0.12 m. Plugging in the values:

x = 0.12 m * cos(2π * 0.50 Hz * 1.0 s) = 0.12 m * cos(π) = -0.12 m,

Therefore, 1.0 second after its release, the glider will be 0.12 m to the left of its equilibrium position.

To find the velocity at this point, we can use the equation:

v = -2πAf * sin(2πft),

where v is the velocity. Plugging in the values:

v = -2π * 0.12 m * 0.50 Hz * sin(2π * 0.50 Hz * 1.0 s) = -0.75 m/s.

Therefore, the glider will have a velocity of -0.75 m/s (to the left) 1.0 second after its release.