A spring of force constant K oscillates with frequency f when suspended with small mass m. What will be the new frequency if the mass suspended is doubled

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Solution

Dear Student ,

The spring constant is inversely proportional to the length.
time period of oscillations
$T = 2 π \sqrt{\frac{m}{k}}$
now mass suspended is doubled than
$T = 2 π \sqrt{\frac{2 m}{k}}$
$n o w t h e r a t i o o f t h e t i m e p e r i o d T_{1} : T_{2} = 1 : \sqrt{2} w e k n o w t h a t t i m e p e r i o d i s i n v e r s a l l y p r o p o r t i o n a l t o f r e q u e n c y f_{1} : f_{2} = \sqrt{2} : 1 s o n e w f r e q u e n c y w i l l b e \frac{f_{1}}{f_{2}} = \sqrt{2} f_{2} = \frac{f_{1}}{\sqrt{2}} R e g a r d s$

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