Asked by James
Could someone show me how to work this? It's for a study guide I'm practicing, and I just can't seem to get the correct answer.
The position of a 0.63 kg object attached to a spring is described by [x = (1.09 m) cos(3.11πt)]
a) Find the object’s position at t = 1.09 s. Answer in meters.
b) Find the object’s acceleration at the same time. Answer in m/s2.
c) What is the frequency of oscillations in Hz?
d) What is the period of oscillations in s?
I've tried using d=x*t for the first part, but I think I must be missing something. This is more advanced stuff than I'm used to so I'm not sure how to go about it
The position of a 0.63 kg object attached to a spring is described by [x = (1.09 m) cos(3.11πt)]
a) Find the object’s position at t = 1.09 s. Answer in meters.
b) Find the object’s acceleration at the same time. Answer in m/s2.
c) What is the frequency of oscillations in Hz?
d) What is the period of oscillations in s?
I've tried using d=x*t for the first part, but I think I must be missing something. This is more advanced stuff than I'm used to so I'm not sure how to go about it
Answers
Answered by
GanonTEK
For part a)
Just substitute in t into the equation you have.
x = (1.09)cos(3.11πt)
x is the displacement and this will come out as an answer in meters. 1.09m is your max amplitute (the number in front of the cos). If you get a negative answer that just means the spring is being compressed so you get a negative displacement
for part b)
I think for this because you have x in terms of t, dx/dt = velocity and dv/dt = acceleration. So if you get d2x/dt2 (the second derivitive) of
(1.09)cos(3.11πt)
and then substitute in the value of t you have for part a) you'll get the acceleration
for part c)
For SHM x = (A)cos(2πt/T)
where t = time and T = period
put 2πt/T = 3.11πt and solve for T
for part d) Frequency = one over the period or
f = 1/T
Just substitute in t into the equation you have.
x = (1.09)cos(3.11πt)
x is the displacement and this will come out as an answer in meters. 1.09m is your max amplitute (the number in front of the cos). If you get a negative answer that just means the spring is being compressed so you get a negative displacement
for part b)
I think for this because you have x in terms of t, dx/dt = velocity and dv/dt = acceleration. So if you get d2x/dt2 (the second derivitive) of
(1.09)cos(3.11πt)
and then substitute in the value of t you have for part a) you'll get the acceleration
for part c)
For SHM x = (A)cos(2πt/T)
where t = time and T = period
put 2πt/T = 3.11πt and solve for T
for part d) Frequency = one over the period or
f = 1/T
Answered by
James
I'm not getting the right answer for part b. Could someone help me understand this part better?
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