The mass that must be added is 0.628 kg
Explanation:
The period of a mass-spring system is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where
m is the mass
k is the spring constant
For the initial mass-spring system in the problem, we have
m = 0.500 kg
T = 1.36 s
Solving for k, we find the spring constant:
[tex]k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m[/tex]
In the second part, we want the period of the same system to be
T = 2.04 s
Therefore, the mass on the spring in this case must be
[tex]m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg[/tex]
Therefore, the mass that must be added is
[tex]\Delta m = 1.128 - 0.500 = 0.628 kg[/tex]
Learn more about period:
brainly.com/question/5438962
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