The end point of a spring oscillates witha period of 2.0s when a block with mass m is attached to it?

Question

When this mass is increased by 2.0kg, the period is found to be 3.0s. Find m.

Answer 1

the period of a spring depends on the sqrt of mass/spring cst

if the period increases by 1.5, then the ratio of mass/k increases by 2.25, so your ratio equation is:

2.25=(m+2)/m so that the original mass is 8/5 kg

Answer 2

the optimal displacement is 3z, the amplitude of the function. From time 0 to time one million, the argument of the cosine has greater by skill of 10w. The cosine is going with the aid of one cycle each 2?. variety of oscillations in one 2nd = 10w/(2?) = 5w/?

Answer 3

T= (2pi) sqrt ( m/k)
m/k = [T(2pi)] ^2

m1/k= [T1(2pi)] ^2
m2/k= [T2(2pi)]^2
k= m2 /[T2(2pi)]^2
Here I have a problem with semantics " increased by 2.0kg" so I shall assume that it was increased to 2.0 kg then
m2= 2kg

then m1=m= k [T1(2pi)] ^2
m= m2 [T1(2pi)/ T2(2pi)]^2
m=2 [2/3]^2= 0.89 kg

Okay if m2 =m+2 kg then

m/m2= k[2pi T1]^2/ k[2pi T2]^2
then
m= m [T1/T2]^2 + 2[T1/R2]^2
m= 2[T1/t2]^2 / (1 - [T1Tt2]^2) or
m= 2/(T2/T1]^2 -1)= 2/([3/2)^2 - 1]
m=1.6kg

Answer 4

(w)=2π/T

(w)²=k/m
(2π/T)²=k/m
(2π/2)²=k/m
k=m(2π/2)²

(2π/3)²=k/m+2

Subsitute k
(2π/3)²=m(2π/2)²/(m+2)
(m+2)4π²/9=m(π)²
4m+8=9m
5m=8
m=1.6 Kg