question_answer

A  light spiral spring supports a 200 g weight at its lower end. It oscillates up and down with a period of 1 s. How much weight (gram) must be removed from the lower end to reduce the period to 0.5 s ?

A)  200                                       

B)  50

C)  53                                         

D)  100

Correct Answer: B

Solution :

                 Let k be the spring constant, then time period (T) of spring-mass system is given by                                 \[T=2\pi \sqrt{\frac{m}{k}}\].                 where m is mass. \[\therefore \]  \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\] Given,   \[{{m}_{1}}=200g,\,\,{{T}_{1}}=1\,s,\,\,{{T}_{2}}-0.5\,s\] \[\Rightarrow \]               \[{{m}_{2}}={{m}_{1}}{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2}}\] \[\therefore \]  \[{{m}_{2}}=200{{\left( \frac{0.5}{1} \right)}^{2}}\]                 \[=0.25\times 200=50\,g\]