Simple Harmonic Motion, Velocity at a certain time

[indietro]

Homework Statement

The velocity of an object in simple harmonic motion is given by v(t)= -(0.347 m/s)sin(15.0t + 2.00π), where t is in seconds. What is the first time after t=0.00 s at which the velocity is -0.100 m/s?

Homework Equations

The Attempt at a Solution

-0.1 = -(0.347)sin(15t + 2\pi)
0.2882 = sin(15t + 2\pi)
15t = arcsin(0.2882) - 2\pi
t = -0.39939 s

- its a neg time so i thought just adding 2\pi would take me to the same point on graph but the positive side : 0.4284 s <-- still the wrong answer

so I am not too sure where I am going wrong.. maybe its just my algebra or is it the whole approach to the question?
thanks!

 

[indietro]

can some one just confirm my algebra, because I am pretty sure it is just a calculation error...? greatly appreciated thanks!

 

[duo]

the sine function is periodic! what does that say about the 2 pi?

0.2882 is correct.

 

[paul232]

1) you say that its v(t) = -vmax sin(15t+2π) Why this 2π becomes 2^π?
2) sin(0.297)= 0.288

So 15t+2π= 2kπ + 0.297 or
15t + 2π = 2kπ + π - 0.297

3) I hope I am right about the numbers.. Your teacher REALLY hates you..

 

[indietro]

so to get the right answer i had to add the negative value i got for the time to the period.
T = 2\pi / \omega = 0.41888

T + t = 0.0195 s ** and this is the right answer

But can someone explain why I had to add it to one cycle (one period) ??

 

[paul232]

Look inside your sine. It is 15t+2p. That means that for t=0 you have sin2p. That implies the body has already finished one period of motion which is not true ( or it is but you started measuring time after 1 period).

think that sin(wt+a) (where a is an angle)

when a =2p that means that before you started timing the body has already completed 1 period.
That is the point of the angle there.
Consider an a= 3p/2 that means that the initital position is not the equillibrium so in order to have the results wrt to x=o we needed that angle.

Now Consider why you find the negative time.. 1) sin(15t+2p)=sin15t ( so you didnt have to evenr write 2p
2) I can't understand why you use arcsin. Also when you have sin(15t+2p)= sin(0.297) <=>
15t+2p= 2kp + 0.297 or 15t+2p=2kp+p-0.297.

3) you need to add a period because you started the timing a period after it started moving. And even if you didnt the +2p implies it. It s better to get rid of it ( because you can ) and solve for t and the same result will pop up.

Hope didnt confuse you more.

 

[Redbelly98]

Here is a simple example. Suppose you need to solve:

sin(x) = 0.5​

So of course we take the arcsin,

x = arcsin(0.5) = π/6 (i.e. 30°)​

And, of course, we can add integer multiples of 2π to get all possible solutions.

But wait a minute ... 5π/6 (i.e. 150°) has a sine of 0.5 also, so that is another solution to the equation. But 5π/6 does not come from adding 2π to π/6, does it?

So where does the extra solution 5π/6 come from? It is because there are two distinct angles that have a sine of 0.5.

 

[AakashR]

v(t)= -(0.347 m/s)sin(15.0t + 2.00π) is equal to v(t)= -(0.347 m/s)sin(15.0t). Now attempt to solve it in the same way as before and you will get t=0.0195s which is the correct answer.