A 0.500kg object attached to a spring with force constant of 8.00n/m vibrates in simple harmonic motion with an amplitude of 10.0cm .calculate a)frequency,b)angular frequency,c)time period,t)the maximum value of its speed and acceleration
Answers
Answer:ω=
m
k
=
0.500 kg
8.00 N/m
=4.00 s
−1
. Assuming the position of the object is at the origin at t=0, position is given by x=10.0sin(4.00t), where x is in cm. From this, we find that v=40.0cos(4.00t), where v is in cm/s, and a=−160sin(4.00t), where a is in cm/s
2
.
(a) v
max
=ωA=(4.00 rad/s)(10.0 cm)=
40.0 cm/s
(b) a
max
=ω
2
A=(4.00 rad/s)
2
(10.0 cm)=
160 cm/s
2
t=0.161 s.
We find then that at that time:
(c) v=(40.0 cm/s)cos[(4.00 Hz)(0.161 s)]=
32.0 cm/s
and
(d) a=−(160 cm/s
2
)sin[(4.00 Hz)(0.161 s)]=
−96.0 cm/s
2
(e) Using t=(
4.00 Hz
1
)sin
−1
(
10.0 cm
x
) we find that when x=0,
t=0, and when x=8.00 cm, t=0.232 s. Therefore, Δt=
0.232 s
Explanation: