1) 2.17 s
The period of a mass-spring oscillating system is given by
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
where k is the spring constant and m is the mass attached to the spring. In this problem, we have
k = 25.1 N/m
m = 3.0 kg
Substituting into the equation, we find
[tex]T=2 \pi \sqrt{\frac{3.0 kg}{25.1 N/m}}=2.17 s[/tex]
2) 0.46 Hz
The frequency of the oscillating system is equal to the reciprocal of the period:
[tex]f=\frac{1}{T}[/tex]
Therefore, by substituting T=2.17 s, we find:
[tex]f=\frac{1}{T}=\frac{1}{2.17 s}=0.46 Hz[/tex]
3) 0.13 s
As before, the period of a mass-spring oscillating system is given by
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
where k is the spring constant and m is the mass attached to the spring. In this part of the problem, we have
k = 25.1 N/m
m = 11 g = 0.011 kg
Substituting into the equation, we find
[tex]T=2 \pi \sqrt{\frac{0.011 kg}{25.1 N/m}}=0.13 s[/tex]
4) 7.69 Hz
The frequency of the oscillating system is equal to the reciprocal of the period:
[tex]f=\frac{1}{T}[/tex]
Therefore, by substituting T=0.13 s, we find:
[tex]f=\frac{1}{T}=\frac{1}{0.13 s}=7.69 Hz[/tex]
5) 1.59 s
Again, the formula for the period of a mass-spring oscillating system is given by
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
In this part of the problem, we have
k = 25.1 N/m
m = 1.6 kg
Substituting into the equation, we find
[tex]T=2 \pi \sqrt{\frac{1.6 kg}{25.1 N/m}}=1.59 s[/tex]
