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Calculus/need help on sinusoidal functions

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Question

Hey Sir, I have a question on which I need your help, If you could please take a look at it..i would greatly appreciate that.

A weight attached to the end of a long spring is bouncing up and down. As it bounces, its distance from the floor varies sinusoidally with time. You start a stopwatch. When the stopwatch reads 0.3 sec, the weight first reaches a high point 60cm above the floor. The next low point, 40 cm above the floor, occurs at 1.8 secs.
a. sketch a graph of this sinusoidal function.
b. write an equation expressing distance from the floor in terms of the number of seconds the stopwatch reads.
c. predict the distance from the floor when the stopwatch reads 17.2 sec.
d. what was the distance from the floor when you started the stopwatch?

Thanks

Answer

drawing
The general sinusoidal function is from the of : B+Asin(ωt). From
the image I attached , you can see that B=50 & A=10. So, our
function will become : h(t)=50+10sin(ωt). All we need to calculate
now is ω. To do so, we claim that tmin-tmax=T/2. Where T is the
period & tmin=1.8sec & tmax=0.3. We also know that ω=2π/T, thus
ω=2π/[2(1.8-0.3)]=2 rad/sec.
As for section c:
17-0.3=16.7
Our 0.3 shifted sin function behaves the same as cos function. Hence,
cos(2*16.7)=-0.211 -> 50+10cos(2*16.7)=47.88
As for section d:
suppose that we activated our stopper x seconds later than the origin, so we may claim that 50+10sin(2*(x+0.3))=60. That gives us
sin(2*(x+0.3))=1 -> 2x+0.6=1 -> 2x=0.4 -> x=0.2 sec.
That means that when we started activating, the weight was
50+10sin(2*0.2)=53.9 cm above the ground.

Alon.

Alon Mandes

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