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Simple harmonic motion and particle speed

  • Thread starter ahhgidaa
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11
0
Motion of particle shows a max acc. of 30 m/s2 and a frequency of 120 cycles per minuet. assuming SHM determine, the amplitude and max velocity.

i found the sqrt of k/m to be 12.57 even though neither was given and the frequency to 2 cycles/sec.

i do not know what relations to use to work out the the rst.

Vmax = Xmax sqrt k/m , but where do i get X max

and where do i find Vo

thank you.just trying to figure it out for studying. apparently i suck at it.
 
Consider that v = dx/dt and a = dv/dt for a point bouncing up and down on the string.
 

Doc Al

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i found the sqrt of k/m to be 12.57 even though neither was given and the frequency to 2 cycles/sec.
To get ω, just convert the given frequency to radians/sec. (Which you already did.)

Vmax = Xmax sqrt k/m , but where do i get X max
That equation is not quite right. (Look it up again.)

Start with a corresponding equation for the maximum acceleration, which will allow you to solve for the amplitude.
 
11
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v=dx/dt= -wAsin(wt)


and x= Acos(wt)

what is the little t, time. is the time the period? and is 2 cycles per second 2 rad/sec. ?

and just fmi, with the differential equation with x(t) when would i use that. when i have as specific time. im just unsure about alot of stuff. sry.
 

Doc Al

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v=dx/dt= -wAsin(wt)


and x= Acos(wt)
Good. Now you're getting somewhere. Continue to get an equation for the acceleration.

what is the little t, time. is the time the period?
t is just the time, not the period.
and is 2 cycles per second 2 rad/sec. ?
No. To convert from cycles to radians, realize that 1 cycle = 2pi radians.

and just fmi, with the differential equation with x(t) when would i use that. when i have as im just unsure about alot of stuff.
Those expressions give you the position and velocity (and, when you get the third equation, acceleration) as a function of time. How would you modify them to get the maximum values of those quantities?
 
11
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well i found the equations on the next page of my book. ha. and have been looking at them and i do not see how they mathematically went from -wAsin(wt) to just wA for Vmax. im looking at the graphs to but my math isnt strong enough to see how they could drop the trig func. but the A i got was .189m and and Vmax of 2.35 m/s. and im glad that im not afraid to ask questions here cuz i never do in class which is my bad. but why isnt it s^2.
 

Doc Al

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44,825
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well i found the equations on the next page of my book. ha. and have been looking at them and i do not see how they mathematically went from -wAsin(wt) to just wA for Vmax. im looking at the graphs to but my math isnt strong enough to see how they could drop the trig func.
What's the maximum value of sine or cosine?

but the A i got was .189m and and Vmax of 2.35 m/s.
Close enough.
but why isnt it s^2.
What do you mean? Why isn't what s^2?
 
11
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for velocity graph the max amplitude is Vmax at 3/4T and for the displacement x at 0 was 3/4T.
how do you get that acc equation? i thinks thats what you were trying to have me do. but what do i think or do to go from V = -wAsin(wt) to Vmax = wA.

dx/dt at x=0? then diff that. which i dnt think i can do.
 

Doc Al

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for velocity graph the max amplitude is Vmax at 3/4T and for the displacement x at 0 was 3/4T.
Again I ask: What's the maximum value that a sine or cosine function can have?
how do you get that acc equation? i thinks thats what you were trying to have me do.
To get the acceleration equation, use a = dv/dt. Take the derivative of the expression you have for V.
but what do i think or do to go from V = -wAsin(wt) to Vmax = wA.
Note that the only variable is in the sin(wt) part. That's why I'm asking you to tell me what's the maximum value for sin(wt).
 
11
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1, -1
 

Doc Al

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Right!

So to get the maximum value in those SHM expressions we can replace cos(ωt) and sin(ωt) by 1. (We just care about the magnitude, so sign isn't important.)

So if y = A cos(ωt), then ymax = A. (And so on.)
 

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