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 of the oscillations, we can use the formula:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

(I) Let's calculate the period:

T = 2π√(0.5000 kg / 4.934 N/m)
T ≈ 2π√(0.1013 s²)
T ≈ 2π * 0.3184 s
T ≈ 2.003 s

So, the period of the oscillations of the mass is approximately 2.003 seconds.

Now, let's find the amplitude of the oscillation.

(II) The amplitude of the oscillation is half the difference between the extreme positions of the pointer.

Amplitude = (70.0 cm - 30.0 cm) / 2
Amplitude = 40.0 cm / 2
Amplitude = 20.0 cm

Therefore, the amplitude of the oscillation is 20.0 cm.

To find the period of the oscillations, we need to use the formula for the period of a mass-spring system:

T = 2π√(m/k)

Where T is the period, m is the mass, and k is the spring constant.

In this case, the mass is m = 0.5000 kg and the spring constant is k = 4.934 N/m. Plugging these values into the formula:

T = 2π√(0.5000 kg / 4.934 N/m)

T = 2π√(0.10126582 kg·m⁻¹)

Calculating this expression:

T ≈ 2π * 0.31853753734 s

Therefore, the period of the oscillations of the mass is approximately 2.0049 seconds.

To find the amplitude of the oscillation, we can use the distance between the equilibrium point (50.0 cm) and one of the extreme positions (30.0 cm or 70.0 cm). The amplitude is half of this distance.

Amplitude = (70.0 cm - 50.0 cm) / 2

Amplitude = 10.0 cm / 2

Amplitude = 5.0 cm

So, the amplitude of the oscillation is 5.0 cm.