Question

In case 1, a block hanging on a spring oscillates with amplitude dd. in case 2, an identical block hanging on an identical spring oscillates with amplitude 2d2d. 1) compare the period of the oscillation in case 1 to the period of the oscillation in case 2.

Answer 1

Answer:

period in case 2 is $\sqrt{2}$ times the period in case 1

Explanation:

The period of oscillation of a spring is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$

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:

$mg = kA$

So we find

$\frac{m}{k}=\frac{A}{g}$

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:

$\frac{m}{k}=\frac{d}{g}$

$T_1=2\pi \sqrt{\frac{d}{g}}$

- for case 2:

$\frac{m}{k}=\frac{2d}{g}$

$T_2=2\pi \sqrt{\frac{2d}{g}}$

And by comparing the two periods, we find:

$\frac{T_2}{T_1}= \sqrt{2}$

So, the period of oscillation in case 2 is $\sqrt{2}$ times the period of oscillation in case 1.