Question

A certain spring-mass system (pictured above) is oscillating up and down when a $0.30kg$ mouse hops on top of the mass to take a ride.
If the original mass is $0.10kg$, how does the new period of the spring-mass system (including the mouse) compare to the old system without the mouse?

• A. The period of the new system is four times as great as the period of the old system
• B. The period of the new system is one-fouth times as great as the period of the old system
• C. The period of the new system is twice as great as the period of the old system
• D. The period of the new system is one-half as great as the period of the old system
• E. The period of the new system is about the same as the period of the old system

Answer 1

The correct option is C The period of the new system is twice as great as the period of the old system

Time-period $T$ of a spring-mass system is given by ,

$T=2\pi \sqrt{\frac{m}{k}}$ ,

where $m=$ mass suspended from spring ,

$k=$ spring constant ,

Now time period of the given system ,

$T=2\pi \sqrt{\frac{0.10}{k}}$ , ...............eq1

Time period of the given new system , when mouse is there ,

$T^{\prime }=2\pi \sqrt{\frac{0.10+0.30}{k}}$ , ................eq2 ,

Therefore $T^{\prime }/T=\sqrt{\frac{0.10+0.30}{0.10}}=\sqrt{4}$ ,

or $T^{\prime }=2T$ ,

The period of new system is twice as great as the period of old system .