What Is the Minimum Time for SHM Particle to Travel Between Two Points?

[gracy]

Homework Statement

A particle performs S.H.M. with amplitude 25 cm and period 3 s. The minimum time required for it to move between two points 12.5 cm on either side of the mean position is ?

Homework Equations

$y$=$a$ $sin$$w$$t$

The Attempt at a Solution

Solution is
$y$=a sin $w$ t

6.25=25 sin$\frac{2π}{3}$t

t=0.25

I don't understand why would we take y=6.25?
I don't understand what " the minimum time required for it to move between two points 12.5 cm on either side of the mean position is " it means?
I take it as total time required to go from +12.5 to mean position and then from mean position to -12.5 cm.

it will take $\frac{T}{4}$ time to reach +a from mean position i.e to reach 25 cm from mean psition
then it will take T/8 time to reach mean position from +12.5 cm . And it will take another T/8 time to reach -12.5cm from mean position hence total time T/8 + T/8 = 2T/8 = 0.25 T
Right?

 

[blue_leaf77]

between two points 12.5 cm on either side

Just want to clarify, by 12.5 cm, is it the distance between the two points, or the distance between each point from the mean position?

 

[gracy]

I don't know. I don't have any extra information . What do you think?

 

[blue_leaf77]

What do you think?

The reason I asked you was because I don't know either. What about the solution you provided, is this part:

Solution is
$y$=a sin $w$ t

6.25=25 sin$\frac{2π}{3}$t

t=0.25

all from the one who gave you the problem? The solution for t=0.25 does not satisfy that equation.

 

[gracy]

I was practicing questions based on SHM. I found the following test
http://career.webindia123.com/career/entrance/ques.asp?sub=physics&groupID=9&course_name=&page=2
Look at question number 7.
Then I started searching for it's solution and I found this
http://www.raoiit.com/MH-CET-2015-Sol/Medical-UG%20MH-CET%202015%20Question%20Paper%20version-11.pdf
Jump to question number 5
Then I looked for it's solution and finally I got
http://www.raoiit.com/MH-CET-2015-Sol/MH-CET-%202015-Answer%20Key%20and%20Solutions%20version-11.pdf

See question number 5.

 

[blue_leaf77]

For some reason I couldn't access the last two links. Nevertheless I tried to work out the problem the way it should be done; it turns out to be the correct answer, which is different from the first solution t=0.25 s.
So, what about your own attempt? Have you checked if it's correct?

 

[gracy]

You tell me. Am I correct ? that's what my OP is all about.

 

[blue_leaf77]

At the bottom end of the first link in post #5, there is a "submit" button you can use to check your answer, you don't need to answer the rest of the question to see your result of problem no. 7. Your answer is 0.25x3 s = 0.75 s which is not among the choices.

then it will take T/8 time to reach mean position from +12.5 cm .

That's not true for non-linear functions, such as sinusoids.

Imagine your particle starts from the mean position at t=0. Then the displacement equation looks like $y(t) = 25 \sin \omega t$. Find the time at which the particle reaches $y=12.5$ for the first time.

 

[gracy]

Your answer is 0.25 ×3s

I did not multiply it by 3.

 

[blue_leaf77]

Then what does T in

0.25 T

stand for?

 

[gracy]

Sorry!

 

[gracy]

When should I use π=3.14 and when to use π=180 degrees?
I think when we are measuring angles we should use π=180 degrees otherwise π=3.14
so in $w$=$\frac{2π}{T}$ π is 180 degrees, right?
But I have seen somewhere frequency of vibrations=f=$\frac{w}{2π}$ and then omega value is given to be 4
f=2/3.14
we can see here π=3.14 in spite of the fact that π is denoting angle.

 

[blue_leaf77]

When should I use π=3.14 and when to use π=180 degrees?

You use 3.14 when the input to the sine function of your calculator is in radians, and use 180 degrees when the input is in degrees. Both values denote an angle, but in different unit.

 

[gracy]

But I have noticed that when π is with sin , cos functions we take it to be 180 degrees and when it isn't accompanied by any such functions it is taken to be 3.14. Why?

 

[cnh1995]

But I have noticed that when π is with sin , cos functions we take it to be 180 degrees and when it isn't accompanied by any such functions it is taken to be 3.14. Why?

In calculus, all the angles are in radians. So π is always 3.14 in calculus. For example, when we write y=sin(2πft) and dy/dt=(2πf)cos(2πft), π is 3.14. But when we express some sinusoidal quantity in this way, y=sin(ωt+∅), ∅ can be written in degrees for our understanding. For example, we write
y=sin(ωt) and y₁=sin(ωt+π) to show that the signals have 180° (or π radian) phase difference. Here, π represents phase difference and can be read as 180°(for our understanding). But when it comes to differentiation or integration, π is always 3.14.

 

[gracy]

Find the time at which the particle reaches $y$=12.5 for the first time

$y(t)$=$25$ sin $w$ $t$

12.5 =$25$ sin $\frac{2π}{3}$ $t$

=$25$ sin $\frac{360}{3}$ $t$

=$25$ sin 120 $t$

=$25$ 0.86 $t$

$\frac{12.5}{25×0.86}$ =$t$

0.58=$t$

 

[blue_leaf77]

No no no, you totally messed up there. The variable $t$ is included in the argument for the sine function: $\sin (\omega t)$.

 

[gracy]

12.5= $25$ sin (120 t)
0.5 = sin (120 t)
(120 t) = 30
t=30/120
t=0.25

 

[blue_leaf77]

12.5= $25$ sin (120 t)
0.5 = sin (120 t)
(120 t) = 30
t=30/120
t=0.25

Yes. Now you just need to find the time needed for the particle to travel from +12.5 to mean position and then from mean position to -12.5 cm.

 

[gracy]

I don't have to include time taken to travel from mean position to 12.5 i.e time I just calculated.

 

[blue_leaf77]

I don't have to include time taken to travel from mean position to 12.5 i.e time I just calculated.

You do, the time $t$ you just found will be used to answer the question. Just use the picture you posted in post#1. What you have now is the travel time from the mean position to +12.5 cm. What is the travel time between the same points for the reverse direction?

 

[gracy]

Yes. I knew it. It will be still 0.25.

 

[blue_leaf77]

Yes. I knew it. It will be still 0.25.

Yes, and the travel time from the mean position to -12.5 cm is?

 

[gracy]

Then I need to find time taken to travel from mean position to -12.5. I think it will be again 0.25. Therefore total time would be 0.25 + 0.25 = 0.5 s

 

[blue_leaf77]

Therefore total time would be 0.25 + 0.25 = 0.5 s

Yes that's the correct answer.

 

[gracy]

So we will not include - negative sign of 12.5 because time is never negative?

 

[blue_leaf77]

Negative time is interpreted as a backward movement in time, we don't need to resort to that concept in this problem.

 

[gracy]

And should we always consider $w$ and t together ? Sin of omega multiplied by t?

 

[blue_leaf77]

And should we always consider www and t together ?

That's a natural combination for an argument of a trigonometric function. Generally though, you have to look at the context. Here it should be clear that a product $\omega t$ where $omega$ is an angular frequency and $t$ is time will give a value in radians. A value in radians (or degree) can become the argument of trigonometric function.

 

[gracy]

Because trig functions have always angles (either in radians or in degrees) as an argument / input . Right? What other quantities can become argument of trig functions?