A 0.12kg body undergoes SHM of amplitude 8.5cm and period 0.20s .
(a) what is the magnitude of the maximum force acting on it .
(b) if the oscillations are produced by a spring,what is the spring constant?
Answers
Explanation:
Answer
(a) The acceleration amplitude is related to the maximum force by Newton’s second law: F
max
=ma
m
. The acceleration amplitude is a
m
=ω
2
x
m
, where ω is the angular frequency (ω=2πf since there are 2π radians in one cycle).
The frequency is the reciprocal of the period: f=
T
1
=
0.20
1
=5.0Hz, so the angular frequency is ω=10π.
Therefore, F
max
=mω
2
x
m
=(0.12kg)(10πrad/s)
2
(0.085m) =10N.
(b) Using Eq. ω=
m
k
, we obtain
k=mω
2
=(0.12kg)(10πrad/s)
2
=1.21×10
2
N/m.
Answer:
(a) The acceleration amplitude is related to the maximum force by Newton’s second law: Fmax=mam. The acceleration amplitude is am=ω2xm, where ω is the angular frequency (ω=2πf since there are 2π radians in one cycle).
The frequency is the reciprocal of the period: f=T1 =0.201=5.0Hz, so the angular frequency is ω=10π.
Therefore, Fmax=mω2xm =(0.12kg)(10πrad/s)2(0.085m) =10N.
(b) Using Eq. ω=mk
, we obtain
k=mω2 =(0.12kg)(10πrad/s)2 =1.21×102N/m.
Explanation: