GS-2 Physical Sciences Question 44

[param_8794381]

This question is wrong.. the passage info itself is wrong... the effective mass must be m1*m2/(m1+m2).. the unit must be a *mass unit*.. The way it is stated in the passage it is inverse mass... inappropriate I would assume.. the spring constant has effective units of kg/second^2. So to cancel the kilograms, kg must be at bottom... for that the formula for effective mass has to change

[jellywing_2058]

Here is how you do this:

v= (1/2π) x √(k/M) where M = (m1+m2/m1*m2)
v= (1/2π) x √[k/(m1+m2/m1*m2)]
v= (1/2π) x √[(k*m1*m2)/(m1+m2]

When each m is doubled:
v= (1/2π) x √[(k*2m1*2m2)/(2m1+2m2)]
v= (1/2π) x √[4k*m1*m2)/(2(m1+m2)]
v= (1/2π) x √[2k*m1*m2)/(m1+m2)]

So v= (1/2π) x √[2k*m1*m2)/(m1+m2)] is bigger than v= (1/2π) x √[k/(m1+m2/m1*m2)] by a factor of √2.

[param_8794381]

I know how to sub in simple equations. I am saying the question is wrong. The formula in the passage gives you a unit of 1/kg for effective mass... mass is never in 1/kg units

[hua8986059]