Pendulums & simple harmonic motion.

[Yehia11]

If we have two pendulums (one with periodic time 1.0 s and the other 0.5 s) and the two are set off oscillating in step, how can we find out the number of times they will be in step over a period of 60 s??

Help much appreciated! thanks!

 

[Ketman]

Well, it looks like one is swinging twice as often as the other. So... what do you conclude?

 

[Yehia11]

Well, it looks like one is swinging twice as often as the other. So... what do you conclude?

Hmm. I'd say, since one is double as quick, that they cross each other every second... that seems to be wrong as the actual answer is apparently 68 times?! :S

 

[torquil]

68 doesn't seem right. What does "in step" mean? I interpreted it as being at the same position, at the same time as moving in the same direction.

Torquil

 

[Ketman]

Have you given the wording of the question exactly as it is in the book or paper?

 

[darpan]

wat do u conclude ketman?

 

[Yehia11]

Yes i have given the wording correctly, but the answer is actually 60. sorry not 68!

it is 60. i drew a sinusoidal graph for both and concluded that they are in step 60 times in 60 second, because they each are at the starting position every second --> once a second!

 

[Yehia11]

But here's another question now where my answer and the actual answer arent equal:

we have a SPRING moving in simple harmonic motion. The amplitude is 40 mm (or 0.04 m)
the frequency is 0.5 Hz so the periodic time is 2 s.

at t=0 the displacement=0 (so we start time when the spring is at equilibrium and is moving upwards)

surely at 1 second the displacement = 0 mm so at 0.5 seconds displacement =40 mm (maximum displacement)

SO at 0.25 seconds the displacement = 20 mm RIGHT? then why is the real answer 28 mm?

Help very much appreciated.

 

[Jebus_Chris]

The graph of position vs time is sinusoidal, not linear.
$$x(t)= A sin \omega t$$
Plug in $$t = .25 s$$ and you get $$x = 28mm$$.

 

[Yehia11]

The graph of position vs time is sinusoidal, not linear.
$$x(t)= A sin \omega t$$
Plug in $$t = .25 s$$ and you get $$x = 28mm$$.

AAAAhhhh yes i remember that equation! hadn't even crossed my mind until i checked my notes. thankyou!

but I still do not see the logic why it is not 20 (because theoretically shouldn't it be half of 40mm?)

And one last question:

If we have 2 simple pendulums, (one longer than the other) oscillating in SHM in step. The next time they are in step is after 20 seconds has elapsed, during which the time the longer pendulum has completed exactly 10 oscillations. Find the string lengths

I found that the length of the short one is 1 metre. but how the other one?? :) thankyou in advance. help very appreciated!

 

[Ketman]

but I still do not see the logic why it is not 20 (because theoretically shouldn't it be half of 40mm?)

You go half the distance in half the time only if the velocity is constant. With harmonic motion you have acceleration and deceleration.

If we have 2 simple pendulums, (one longer than the other) oscillating in SHM in step. The next time they are in step is after 20 seconds has elapsed, during which the time the longer pendulum has completed exactly 10 oscillations. Find the string lengths

I found that the length of the short one is 1 metre. but how the other one?? :) thankyou in advance. help very appreciated!

That was a big jump! The shorter one is the one you know least about to begin with. How did you arrive at your answer?

 

[Yehia11]

You go half the distance in half the time only if the velocity is constant. With harmonic motion you have acceleration and deceleration.

Ah right, yea i see it now, thanks!

That was a big jump! The shorter one is the one you know least about to begin with. How did you arrive at your answer?

hahaha sorry my bad, i DONT KNOW the length of the shorter one! that's what I am trying to find. i found out that the long one's length is one metre through this formula:

T (periodic time) = 2pi√(l/g) and so the long one is 1 metre, I want to find the short one.

Can you help?

 

[Ketman]

Well, you might need someone smarter than me. It seems clear that the shorter pendulum will be in step with the longer one after 10 oscillations if it goes just 10% faster. But that also applies if it's 110% faster, or 210%, or 310%... Or so it seems to me. But since there can only be one answer, it looks like I must be wrong.

 

[Jebus_Chris]

So you know the length and period of the large pendulum, 2s and 1m respectively. The position of the large pendulum after 20 seconds is the same as where it began (x=A). The smaller pendulum must go through some amount of oscillations in 20 seconds. We know that the period will be shorter than the other period so use the previous equation I gave you and plug in final position and time and find T.

 

[Redbelly98]

Moderator's note: in the future, separate questions should be posted in a new thread.