Two ingenious 120 student students decide to use an oscillating mass on a spring to check the shutter (exposure) time of their camera. They set up vertically alongside the mass and spring. A pointer attached to the mass allows the scale on the meter stick to be read as illustrated in the diagram. The spring has a constant of k=4.934 N/m, and the mass is m=0.5000 kg. In equilibrium, the pointer is at the 50.0cm mark of the meter stick. When the mass is oscillating, the pointer movies such that at its extreme positions it falls exactly on the 30.0cm and 70.0cm scale marks.
(I) What is the period of the of the oscillations of the mass?
(II) What is the amplitude of the oscillation?
____________________
| | |
70cm | | \spring
| | /_
50cm | | |__|mass
| |
30cm |_|
/\
| Meter stick
To find the period (T) of the oscillations of the mass, we can use the equation for the period of a mass-spring system:
T = 2π * √(m/k)
where m is the mass and k is the spring constant.
In this case, m = 0.5000 kg and k = 4.934 N/m.
Plugging these values into the equation, we get:
T = 2π * √(0.5000 kg / 4.934 N/m)
T = 2π * √(0.1013 kg/m)
T ≈ 2π * 0.3186 s
T ≈ 2.00 s
So, the period of the oscillations is approximately 2.00 seconds.
To find the amplitude of the oscillation, we can use the distance between the extreme positions of the pointer. In this case, the distance between the 30.0 cm and 70.0 cm scale marks is 70.0 cm - 30.0 cm = 40.0 cm.
However, since the displacement measured by the pointer is from the equilibrium position (50.0 cm mark), we need to subtract the displacement at equilibrium from the total distance:
Amplitude = (40.0 cm - 50.0 cm) / 2
Amplitude = -5.0 cm / 2
Amplitude = -2.5 cm
Therefore, the amplitude of the oscillation is -2.5 cm.