Question

A certain spring mass system (pictured below) 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 0.10 kg. 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-fourth 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.

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

$T=2\pi \sqrt{m/k}$

where $m=$ mass of spring , $k=$ spring constant

given $m=0.10kg$ ,

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

now when the mouse is in the mass, then mass becomes $m^{\prime }=0.10+0.30=0.40kg$ '

therefore new time period will be ,

$T^{\prime }=2\pi \sqrt{0.40/k}$ ..........................eq2

dividing eq2 by eq1 , we get

$T^{\prime }/T=\sqrt{0.40/0.10}=\sqrt{4}$ ,

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

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