Respuesta :
Answer:
389 kg
Explanation:
The computation of mass is shown below:-
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
Where K indicates spring constant
m indicates mass
For the new time period
[tex]T^' = 2\pi \sqrt{\frac{m'}{k} }[/tex]
Now, we will take 2 ratios of the time period
[tex]\frac{T}{T'} = \sqrt{\frac{m}{m'} }[/tex]
[tex]\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]0.5625 = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]m' = \frac{0.500}{0.5625}[/tex]
= 0.889 kg
Since mass to be sum that is
= 0.889 - 0.500
0.389 kg
or
= 389 kg
Therefore for computing the mass we simply applied the above formula.
The mass added to the object to change the period to 2.00 s is 0.389 kg and this can be determined by using the formula of the time period.
Given :
A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s.
The formula of the time period is given by:
[tex]\rm T = 2\pi\sqrt{\dfrac{m}{K}}[/tex] ---- (1)
where m is the mass and K is the spring constant.
The new time period is given by:
[tex]\rm T'=2\pi\sqrt{\dfrac{m'}{K}}[/tex] ---- (2)
where m' is the total mass after the addition and K is the spring constant.
Now, divide equation (1) by equation (2).
[tex]\rm \dfrac{T}{T'}=\sqrt{\dfrac{m}{m'}}[/tex]
Now, substitute the known terms in the above expression.
[tex]\rm \dfrac{1.50}{2}=\sqrt{\dfrac{0.5}{m'}}[/tex]
Simplify the above expression in order to determine the value of m'.
[tex]\rm m'=\dfrac{0.5}{0.5625}[/tex]
m' = 0.889 Kg
Now, the mass added to the object to change the period to 2.00 s is given by:
m" = 0.889 - 0.500
m" = 0.389 Kg
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https://brainly.com/question/2144584
