the length of a spring increases by 0.25m when the body of mass 0.6kg is suspended from it. a body of mass 0.24kg is suspended and stretched downwards and left. then what will be the value of time period of spring
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Answered by
2
Answer: T=2π
k
m
=2π
25
0.25
=
Explanation:
The force constant of the spring,
k=
x
F
=
0.2
0.5×10
=25N/m
Now, if the mass of 0.25 kg is suspended by the spring, then the period of oscillation,
T=2π
k
m
=2π
25
0.25
=
Answered by
0
Explanation:
the answer is 55. which meants its a prime number
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