Simple Harmonic Motion of skydiver

[BDR]

I can't seem to get the right answer, where is my mistake?

A 90 kg skydiver hanging from a parachute bounces up and down with a period of 1.5 seconds. What is the new period of oscillation when a second skydiver, whose mass is 60 kg, hangs from the legs of the first?

I am using the equation T = 2pie square root of m/k.

I found k and then combined the 2 masses (m1+m2), and put that back into the equation and found an answer but not the right one.

The right answer is suppose to be 1.94 s ...i'm getting 58

 

[LeonhardEuler]

Your method is correct, you must be doing some calculation wrong. What did you get for k?

 

[BDR]

I got k = 38.14 I don't know if its right though

 

[LeonhardEuler]

There's your problem. k is much bigger than that. How did you find k?

 

[mukundpa]

your value of k is not correct check the calculations

You may also use T1/T2 = squrt of m1/(m1+m2)

 

[BDR]

I'm not sure that is the correct number i got for K, i have so many numbers on my paper everything is mixed together. I took the period given which was 1.5 divided by 2 pie 1.5/2pie = .239. Then i squared to get rid of the square root. Which gave me .239 = m/k. Then multiplied by k and divided.

 

[BDR]

I'm getting an answer of 2.5, is that correct? The correct answer is 1.94 s, or that's what the book says.

 

[mukundpa]

T = 2p sqr(m/K)
T^2= 4p^2(m/K)
K = (4*P^2*m)/T^2
put the values it gives 1579.14 N/m
how you were calculating

I suggested an easy method in the earlier posting
As K is not changing you can eliminate it from the two cases by dividing the equations of the two time periods you will get the equation I gave in that posting and by substituting the values you can calculate T2

It is always better to solve first in notations and finally calculating the numerical value.