Answer:
period in case 2 is
times the period in case 1
Explanation:
The period of oscillation of a spring is given by:
![T=2 \pi \sqrt{\frac{m}{k}}](https://tex.z-dn.net/?f=T%3D2%20%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%7D)
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:
![mg = kA](https://tex.z-dn.net/?f=mg%20%3D%20kA)
So we find
![\frac{m}{k}=\frac{A}{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bk%7D%3D%5Cfrac%7BA%7D%7Bg%7D)
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:
![\frac{m}{k}=\frac{d}{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bk%7D%3D%5Cfrac%7Bd%7D%7Bg%7D)
![T_1=2\pi \sqrt{\frac{d}{g}}](https://tex.z-dn.net/?f=T_1%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7Bd%7D%7Bg%7D%7D)
- for case 2:
![\frac{m}{k}=\frac{2d}{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bk%7D%3D%5Cfrac%7B2d%7D%7Bg%7D)
![T_2=2\pi \sqrt{\frac{2d}{g}}](https://tex.z-dn.net/?f=T_2%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7B2d%7D%7Bg%7D%7D)
And by comparing the two periods, we find:
![\frac{T_2}{T_1}= \sqrt{2}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7BT_1%7D%3D%20%5Csqrt%7B2%7D)
So, the period of oscillation in case 2 is
times the period of oscillation in case 1.