Finding (C): Halfway Between Equilibrium and End Point

[hidemi]

Homework Statement

A particle is in simple harmonic motion with period T. At time t = 0 it is halfway
between the equilibrium point and an end point of its motion, traveling toward the
end point. The next time it is at the same place is:
A. t =T
B. t =T/2
C. t =T/3
D. t = T/4
E. none of the above

The answer is C

Relevant Equations

X = A cos(wt+c)

I think (D) is correct since it is half way between the equilibrium point and and the end point of its motion, it is a quarter of the total distance. How to get (C)?

 

[anuttarasammyak]

I think t=T is ,though not least, the time it is in the same place. B to D are inappropriate because halfway means anywhere between the top and the bottom.

 

[patric44]

can you find the angle between $t_{1}$ and $t_{2}$

 

[Steve4Physics]

Suppose the motion is between points A and B:

A P O Q B

The centre is O , the end points are A and B.
P is the midpoint of AO.
Q is the midpoint of OB.

Say we want the time for Q→B→Q.

The following times are equal:
$T_{OB} = T_{BO} = T_{OA} = T_{AO} = \frac T 4$
This a due to the inherent symmetry in a cycle of simple harmonic motion. It can be seen by looking at the shape of a sine curve.

But $T_{AP}, T_{PO}$, etc. are not equal to any 'obvious' value (not $\frac T 8$ for example). Look at a sine curve carefully.

With O at x=0, taking x=0 when t=0 and assuming velocity at t=0 is positive, we can describe the motion by:$$x(t) = Asin \left[ \left( \frac {2\pi} T \right) t \right]$$(Using sine rather than cosine is most convenient here.)

Suppose we want $T_{QB}$. (The required ‘return time' will be twice this.)

I think the simplest method to find $T_{QB}$ is first to find $T_{OQ}$ and then subtract it from $T_{OB}$ (since we already know that $T_{OB} =\frac T 4$).

The position of Q is x = $\frac A 2$ which gives:$$\frac A 2 = Asin \left[ \left( \frac {2\pi} T \right) T_{OQ} \right]$$Solve for $T_{OQ}$. And take it from there.

Someone might post a simpler method, but the above method is easier than it might initially appear!

 

[hidemi]

Suppose the motion is between points A and B:

A P O Q B

The centre is O , the end points are A and B.
P is the midpoint of AO.
Q is the midpoint of OB.

Say we want the time for Q→B→Q.

The following times are equal:
$T_{OB} = T_{BO} = T_{OA} = T_{AO} = \frac T 4$
This a due to the inherent symmetry in a cycle of simple harmonic motion. It can be seen by looking at the shape of a sine curve.

But $T_{AP}, T_{PO}$, etc. are not equal to any 'obvious' value (not $\frac T 8$ for example). Look at a sine curve carefully.

With O at x=0, taking x=0 when t=0 and assuming velocity at t=0 is positive, we can describe the motion by:$$x(t) = Asin \left[ \left( \frac {2\pi} T \right) t \right]$$(Using sine rather than cosine is most convenient here.)

Suppose we want $T_{QB}$. (The required ‘return time' will be twice this.)

I think the simplest method to find $T_{QB}$ is first to find $T_{OQ}$ and then subtract it from $T_{OB}$ (since we already know that $T_{OB} =\frac T 4$).

The position of Q is x = $\frac A 2$ which gives:$$\frac A 2 = Asin \left[ \left( \frac {2\pi} T \right) T_{OQ} \right]$$Solve for $T_{OQ}$. And take it from there.

Someone might post a simpler method, but the above method is easier than it might initially appear!

I got it! Thanks so much!