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Mechanical Properties Of Fluids

Question
CBSEENPH11019294

The length of spring of the oscillator is L and beats one second when loaded by mass m. If spring is cut into two equal parts each of length L/2, then by what percentage the loaded mass must be changed so that the time period of oscillation remains unchanged? 

Solution
Time period of spring pendulum is given by,
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If the spring is cut into two equal parts, the spring constant of each part becomes equal to 2k.
Mass of the load is increased to m' to keep the time period same. 
Thus comma space space space
straight T equals 2 straight pi square root of straight m over straight k end root equals 2 straight pi square root of fraction numerator straight m apostrophe over denominator 2 straight k end fraction end root space

rightwards double arrow space space space space space space space space space space space space space straight m to the power of apostrophe space equals space 2 straight m
 
Therefore, the percentage increase in mass is 100%.