An air-track glider attached to a spring oscillates between the 3.00 cm mark and the 50.0 cm mark on the track. The glider completes 10.0 oscillations in 36.0 s. What is the period of the oscillations?
What is the frequency of the oscillations?
What is the angular frequency of the oscillations?
What is the amplitude?
What is the maximum speed of the glider?
T = time/oscillations = 36/10 = 3.6s
f = 1/T = 1/3.6 = .278
Angular f = 2Pi/T = 2Pi/3.6 = 1.74
X (amplitude) = total distance/2 = (50cm - 3cm)/2 = 23.5 cm or .235 m
T = (2PiX)/Vmax or Vmax = (2PiX)/T = (2Pi0.235)/3.6 = .410 m/s
To find the period of the oscillations, you need to know the total time it takes for the glider to complete one full oscillation (back-and-forth motion). In this case, the glider completes 10.0 oscillations in 36.0 s.
To find the period, divide the total time by the number of oscillations:
Period = Total Time / Number of Oscillations
In this case, the total time is 36.0 s and the number of oscillations is 10.0:
Period = 36.0 s / 10.0 = 3.60 s
Therefore, the period of the oscillations is 3.60 s.
To find the frequency of the oscillations, you need to know the number of oscillations per second. The frequency is the reciprocal of the period:
Frequency = 1 / Period
In this case, the period is 3.60 s, so the frequency is:
Frequency = 1 / 3.60 s = 0.2778 Hz
Therefore, the frequency of the oscillations is 0.2778 Hz.
To find the angular frequency of the oscillations, you need to know the frequency in radians per second. The angular frequency is related to the frequency:
Angular Frequency = 2π × Frequency
In this case, the frequency is 0.2778 Hz:
Angular Frequency = 2π × 0.2778 Hz = 1.745 rad/s
Therefore, the angular frequency of the oscillations is 1.745 rad/s.
To find the amplitude of the oscillations, you need to know the maximum displacement (distance) of the glider from its equilibrium position. In this case, the glider oscillates between the 3.00 cm mark and the 50.0 cm mark on the track.
The amplitude is half the total distance traveled by the glider:
Amplitude = (Maximum Displacement) / 2
In this case, the maximum displacement is 50.0 cm - 3.00 cm = 47.0 cm:
Amplitude = 47.0 cm / 2 = 23.5 cm
Therefore, the amplitude of the oscillations is 23.5 cm.
To find the maximum speed of the glider, you need to use the formula for maximum speed in simple harmonic motion:
Maximum Speed = Angular Frequency × Amplitude
In this case, the angular frequency is 1.745 rad/s and the amplitude is 23.5 cm:
Convert the amplitude to meters: 23.5 cm = 0.235 m
Maximum Speed = 1.745 rad/s × 0.235 m = 0.409 m/s
Therefore, the maximum speed of the glider is 0.409 m/s.