Answer:
period in case 2 is [tex]\sqrt{2}[/tex] times the period in case 1
Explanation:
The period of oscillation of a spring is given by:
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
where
m is the mass hanging on the spring
k is the spring constant
Therefore, in order to compare the period of the two springs, we need to find their m/k ratio.
We know that when a mass hang on a spring, the weight of the mass corresponds to the elastic force that stretches the spring by a certain amplitude A:
[tex]mg = kA[/tex]
So we find
[tex]\frac{m}{k}=\frac{A}{g}[/tex]
The problem tells us that the amplitude of case 1 is d, while the amplitude in case 2 is 2d. So we can write:
- for case 1:
[tex]\frac{m}{k}=\frac{d}{g}[/tex]
[tex]T_1=2\pi \sqrt{\frac{d}{g}}[/tex]
- for case 2:
[tex]\frac{m}{k}=\frac{2d}{g}[/tex]
[tex]T_2=2\pi \sqrt{\frac{2d}{g}}[/tex]
And by comparing the two periods, we find:
[tex]\frac{T_2}{T_1}= \sqrt{2}[/tex]
So, the period of oscillation in case 2 is [tex]\sqrt{2}[/tex] times the period of oscillation in case 1.
