A body of mass m tied to a spring perform S.H.M. with period 2 seconds. If the mass is increase by 3m. What will be the period of S.H.M.
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Given info : A body of mass m tied to a spring performs simple harmonic motion with period 2 seconds. if mass is increased by 3m.
To find : the period of simple harmonic motion is..
solution : time period of simple harmonic is given by, T = 2π√{m/K}
where K is spring constant, m is mass of body performing SHM.
here T = 2 seconds
so, 2 = 2π√{m/K}
⇒1/π² = m/K. ....(1)
now mass is increased by 3m
so, new mass of body is , m' = (m + 3m) = 4m
T' = 2π√{m'/K}
= 2π√{4m/K}
from equation (1),
= 2π√{4 × 1/π²}
= 2π × 2/π
= 4 seconds
Therefore the time period of SHM will be 4 seconds.
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