An object suspended from a spring is oscillating up and down. The distance from the high point to the low point is 30cm, and the object takes 4s to complete 5cycles. For the first few cycles, the distance feom the mean position, d(t) centimetres, with respect to the time, t seconds, is modelled by a sine function.
Find the amplitude, vertical shift, phase shift and k value and period. Do all calculations and then make your final equation.
Can u please help me with this pne cuz there are a lot more and i don' t understand any of them please help meeee!!!:(
"distance from the high point to the low point is 30cm"
----> your a value is 15
so far: d(t) = 15 sin ( ???? t)
5 cycles takes 4 seconds
so 1 cycle takes 4/5 seconds
then 2π/k = 4/5
4k = 10π
k = 5π/2
so far: d(t) = 15 sin (5π/2)( t ± phase shift)
since you didn't say where you want the object to be at t-0 , I will assume at the mean positon of zero
so .....
d(t) = 15sin (5π/2)t
since you only had data affecting the amplitude and period I will go with the above.
testing:
since the period is .8 seconds we should have a value of -15 at the 3/4 mark
3/4 (.8) = .6
set your calculator to radians and evaluated
15 sin (5π/2)(.6)
( I get -15)
Im sorry I do not understand. How did you get 15 and the period .8 seconds how did you get the 5pie/2please explainI am so confused!!!:(
I assumed you were familiar with the basic general sine function
y = a sin kØ
where a is the amplitude, and k is such that
2π/k is the period of the curve.
To be able to do the type of question above, you just MUST know that terminology.
Check with your textbook or your notes.
Sure, I'd be happy to help you with this problem!
To find the amplitude, we need to determine the maximum distance from the mean position that the object oscillates. In this case, the distance from the high point to the low point is given as 30 cm. Since the object oscillates symmetrically, the amplitude is half of this distance. So, the amplitude is 30 cm ÷ 2 = 15 cm.
Next, let's find the vertical shift or the mean position. The mean position is the equilibrium point where the object rests when it is not oscillating. In this case, the mean position is the midpoint between the high point and the low point. So, the vertical shift is (30 cm / 2) = 15 cm above the x-axis.
Now, to determine the phase shift, we need to find the time at which the object starts oscillating. The problem states that the object takes 4 seconds to complete 5 cycles. Therefore, the time for one full cycle is 4 seconds ÷ 5 = 0.8 seconds. The phase shift is the horizontal shift of the function, which is related to the time at which the function starts oscillating. Since the object starts at the midpoint and oscillates upward, the phase shift is 0s.
The k value in this case refers to the angular frequency of the oscillation. To find the k value, we need to determine the period of the oscillation. The period is the amount of time it takes for one complete cycle. We already found that the time for one full cycle is 0.8 seconds. Therefore, the period is 0.8 seconds.
The general equation for a sine function is given as:
y = A * sin(k(t - d)) + C
Where:
A represents the amplitude,
k represents the angular frequency,
d represents the phase shift,
and C represents the vertical shift or mean position.
In our case, the equation will be:
d(t) = 15 * sin((2π / 0.8)(t - 0)) + 15
Simplifying the equation, we get:
d(t) = 15 * sin((2.5π)(t)) + 15
Therefore, the final equation is:
d(t) = 15 * sin(2.5πt) + 15
I hope this clears up the confusion and helps you understand the problem better! If you have any more questions, feel free to ask.