Asked by Samantha
A 215 g mass attached to a horizontal spring oscillates at a frequency of 5.50 Hz. At t=0s, the mass is at x= 6.40 cm and has vx =- 45.0 cm/s. Determine:
a) The Maximum acceleration to 3 significant digits.
- I calculated this to be 7.64 but it keeps saying I have the wrong SD and that I rounded something off wrong.
b) The total energy.
c) The position at t= 3.20s.
a) The Maximum acceleration to 3 significant digits.
- I calculated this to be 7.64 but it keeps saying I have the wrong SD and that I rounded something off wrong.
b) The total energy.
c) The position at t= 3.20s.
Answers
Answered by
Damon
I am doing all these in meters and seconds
2 pi f = sqrt (k/m) = 34.6
2 pi (5.5) = sqrt (k/.215)
1194 = k/.215
k = 2568 N/m
x = b sin 34.6 t + c cos 34.6 t
Vx = 34.6 b cos 3.46 t - 34.6 c sin 34.6 t
now put in t = 0
.0640 = 0 b + c so c = .0640
-.45 = 34.6 b - 0
so b = -.013
Therefore finally
x = -.013 sin 34.6 t + .0640 cos 34.6 t
Vx = -.45 cos 34.6 t - 2.21 sin 34.6 t
and then a, the acceleration
a = +15.57 sin 34.6 t -76.5 cos 34.6 t
max a = sqrt (15.57^2+76.5^2 )
max a = sqrt (6095)
max a = 78.1
For total energy
Ke = (1/2)mv^2 + (1/2)kx^2
you can find x at t = 0 from equation above, we were given v and we calculated k
for the position, just put 3.20 in your equation for x
2 pi f = sqrt (k/m) = 34.6
2 pi (5.5) = sqrt (k/.215)
1194 = k/.215
k = 2568 N/m
x = b sin 34.6 t + c cos 34.6 t
Vx = 34.6 b cos 3.46 t - 34.6 c sin 34.6 t
now put in t = 0
.0640 = 0 b + c so c = .0640
-.45 = 34.6 b - 0
so b = -.013
Therefore finally
x = -.013 sin 34.6 t + .0640 cos 34.6 t
Vx = -.45 cos 34.6 t - 2.21 sin 34.6 t
and then a, the acceleration
a = +15.57 sin 34.6 t -76.5 cos 34.6 t
max a = sqrt (15.57^2+76.5^2 )
max a = sqrt (6095)
max a = 78.1
For total energy
Ke = (1/2)mv^2 + (1/2)kx^2
you can find x at t = 0 from equation above, we were given v and we calculated k
for the position, just put 3.20 in your equation for x
Answered by
Damon
Ke = (1/2)mv^2 + (1/2)kx^2
should be Total = Ke+U = (1/2) m v^2 + (1/2) k x^2
should be Total = Ke+U = (1/2) m v^2 + (1/2) k x^2
Answered by
Damon
Check everything I did for arithmetic errors.
Answered by
Samantha
I don't get it. I'm getting x = 0. Is that supposed to be right?
Answered by
Samantha
Also for position I'm getting 0.145 when I use this equation =>
x = -.013 sin 34.6 t + .0640 cos 34.6 t
x = -.013 sin 34.6 t + .0640 cos 34.6 t
Answered by
Damon
You said:
Also for position I'm getting 0.145 when I use this equation =>
x = -.013 sin 34.6 t + .0640 cos 34.6 t
when t = 0, sin 34.6 t = 0 because sin 0 is 0
when t = 0, cos 34.6 t = 1 because cos 0 = 1
so you get
x = -.013 * 0 + .0640 * 1
Also for position I'm getting 0.145 when I use this equation =>
x = -.013 sin 34.6 t + .0640 cos 34.6 t
when t = 0, sin 34.6 t = 0 because sin 0 is 0
when t = 0, cos 34.6 t = 1 because cos 0 = 1
so you get
x = -.013 * 0 + .0640 * 1
Answered by
Damon
x = -.013 sin 34.6 t + .0640 cos 34.6 t
ok, put in t = 3.20 seconds
x = -.013 sin (111) + .0640 cos(111)
that 111 is in radians
so
x = -.013 (-.865) +.0640 (-.503)
= -.0209 meters = 2.09 cm
ok, put in t = 3.20 seconds
x = -.013 sin (111) + .0640 cos(111)
that 111 is in radians
so
x = -.013 (-.865) +.0640 (-.503)
= -.0209 meters = 2.09 cm
Answered by
Samantha
Are you getting 5.28 J for the total energy?
Answered by
Damon
Hang on, I will check
Answered by
Damon
for energy
(1/2) (.215)(-.45)^2 +(1/2)(1194)(.0640)^2
= .0218+2.45 = 2.47 Joules
(1/2) (.215)(-.45)^2 +(1/2)(1194)(.0640)^2
= .0218+2.45 = 2.47 Joules
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