# SHM particle min time

## 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?

Last edited:

blue_leaf77
Homework Helper
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
blue_leaf77
Homework Helper
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 [Broken]
Then I looked for it's solution and finally I got

See question number 5.

Last edited by a moderator:
blue_leaf77
Homework Helper
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
Homework Helper
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
I did not multiply it by 3.

blue_leaf77
Homework Helper
Then what does T in
0.25 T
stand for?

gracy
Sorry!

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
Homework Helper
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
Homework Helper
Gold Member
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
Homework Helper
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
Homework Helper
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
Homework Helper
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
Homework Helper
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
Homework Helper
Therefore total time would be 0.25 + 0.25 = 0.5 s

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

blue_leaf77
Homework Helper
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