An object goes through simple harmonic motion with its position described as

In summary, the object moves through simple harmonic motion with a position described by x(t) = 50sin(2piet/3). At first, v(t) = 50cos(2.094t)(2.094), but then v(t) = 80 when is Velocity first equal to 80.
  • #1
losthelp
10
0
An object goes through simple harmonic motion with its position described as...

Homework Statement


An object undergoes simple harmonic motion with its position described by
x(t ) = 50sin (2piet/3)

C) when is Velocity first equal to 80 D) whe nis the Velocity next equal to 80


Homework Equations




The Attempt at a Solution



k iknow you have to find the derivative.. which i think is this.

V(t)= 50cos(2.094t)(2.094)
= 104.7cos(2.094t)

then set it to 80: 80=104.7sin(2.094t)
i don't know where to go from here :S

and for D) i have no idea wher to start!
 
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  • #2


if anyone could take a stab at it hatd be great :)
 
  • #3


losthelp said:

Homework Statement


An object undergoes simple harmonic motion with its position described by
x(t ) = 50sin (2piet/3)

C) when is Velocity first equal to 80 D) whe nis the Velocity next equal to 80


Homework Equations




The Attempt at a Solution



k iknow you have to find the derivative.. which i think is this.

V(t)= 50cos(2.094t)(2.094)
= 104.7cos(2.094t)

then set it to 80: 80=104.7sin(2.094t)
i don't know where to go from here :S

and for D) i have no idea wher to start!

Don't replace 2pi/3 with an approximate value.
You have x(t) = 50 sin((2pi/3)t)
so, v(t) = x'(t) = 50 cos((2pi/3)t) * (2pi/3) = 100 pi/3 * cos((2pi/3)t).

Now, solve for t at the first time when v(t) = 80. Think about how the cosine graph looks when you answer this question and the other one.
 
  • #4


Mark44 said:
Don't replace 2pi/3 with an approximate value.
You have x(t) = 50 sin((2pi/3)t)
so, v(t) = x'(t) = 50 cos((2pi/3)t) * (2pi/3) = 100 pi/3 * cos((2pi/3)t).

Now, solve for t at the first time when v(t) = 80. Think about how the cosine graph looks when you answer this question and the other one.

why does it matter if i make it 100pi/3 * cos((2pi/3)t) ...isnt it the same thing?

well if i solve it like that: 80= 100pi/3 * cos(2pi/3t)
0.7639= cos((2pi/3)t)
2.2918=cos ((2pi)t)
t= 2.918

and then how ould u solve for it next being 80 :S is that just solving the v(t) equation with time at 2.918?
 
  • #5


mark can you please answer me because I am really quite desparate now!
 
  • #6


losthelp said:
why does it matter if i make it 100pi/3 * cos((2pi/3)t) ...isnt it the same thing?

well if i solve it like that: 80= 100pi/3 * cos(2pi/3t)
0.7639= cos((2pi/3)t)
2.2918=cos ((2pi)t)
t= 2.918

and then how ould u solve for it next being 80 :S is that just solving the v(t) equation with time at 2.918?

104.7 is not the same number as 100 pi/3. They're not too far apart, but they are different from the 2nd decimal place on.

You will always get more exact results in your answers if you save your approximations for the last step. If you do them too early, you are likely to introduce greater and greater errors, making your final answer even more inaccurate.

Here's your work from above:
80= 100pi/3 * cos(2pi/3t)
0.7639= cos((2pi/3)t)
2.2918=cos ((2pi)t)
t= 2.918

Your step from the 2nd equation to the 3rd is wrong. You apparently multiplied by 3 on both sides, but 3 *cos((2pi/3)t) != cos(2pi t)

Let's start again from the first equation:
100pi/3 * cos(2pi/3t) = 80
cos((2pi/3)t) = 240/(100 pi) = 2.4/pi

Now, what can I do to get (2pi/3)t all by itself on the left side?

If I can do that, then I can multiply both sides by 3/(2pi) to get t.

For t, I get .334890728 for the first solution. Even if you hadn't made a mistake, I think that your answer would have agreed with mine in only three or four decimal places.

For the second solution, you need to know the period of your function and what a cosine function's graph looks like.
 
  • #7


Mark44 said:
104.7 is not the same number as 100 pi/3. They're not too far apart, but they are different from the 2nd decimal place on.

You will always get more exact results in your answers if you save your approximations for the last step. If you do them too early, you are likely to introduce greater and greater errors, making your final answer even more inaccurate.

Here's your work from above:
80= 100pi/3 * cos(2pi/3t)
0.7639= cos((2pi/3)t)
2.2918=cos ((2pi)t)
t= 2.918

Your step from the 2nd equation to the 3rd is wrong. You apparently multiplied by 3 on both sides, but 3 *cos((2pi/3)t) != cos(2pi t)

Let's start again from the first equation:
100pi/3 * cos(2pi/3t) = 80
cos((2pi/3)t) = 240/(100 pi) = 2.4/pi

Now, what can I do to get (2pi/3)t all by itself on the left side?

If I can do that, then I can multiply both sides by 3/(2pi) to get t.

For t, I get .334890728 for the first solution. Even if you hadn't made a mistake, I think that your answer would have agreed with mine in only three or four decimal places.

For the second solution, you need to know the period of your function and what a cosine function's graph looks like.

THANK YOU! now the period of my function is 3, and i know how a cosine graph looks like. so could u at all tell me what I am suposed to be looking for :S sorry/thanks!
 
  • #8


The first time v(t) = 80 at [tex]t \approx .334890728 [/tex]. The next time is when [tex]t \approx 3 - .334890728 [/tex].
 
  • #9


thats it? hmmm thank you so much mark! you're a life saver!

XOXOXOOXOXOXOXO
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth around a central point with a constant amplitude and period. It occurs when the restoring force on the object is directly proportional to the displacement from the central point.

2. What factors affect the period of simple harmonic motion?

The period of simple harmonic motion is affected by the mass of the object, the force constant of the spring (if present), and the amplitude of the oscillation. The period is directly proportional to the square root of the mass and inversely proportional to the force constant and amplitude.

3. How is the position of an object in simple harmonic motion described?

The position of an object in simple harmonic motion is typically described using a sine or cosine function, where the amplitude represents the maximum displacement from the central point and the period represents the time it takes for the object to complete one full oscillation.

4. Can an object in simple harmonic motion have a negative position?

Yes, an object in simple harmonic motion can have a negative position. This simply means that the object is on the opposite side of the central point from its starting position. The position of the object can vary between positive and negative values as it oscillates back and forth.

5. What is the relationship between frequency and period in simple harmonic motion?

The frequency and period of simple harmonic motion are inversely proportional. This means that as the frequency increases, the period decreases, and vice versa. The frequency is the number of oscillations per unit time, while the period is the time it takes for one full oscillation.

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