physics question help please oscillations
Hi, please could I have some help on part c on page 8? I'm not sure how I could work out the position of max velocity/acceleration?
Thanks!
paper: https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FPhysics%2FA-level%2FTopic-Qs%2FCAIE%2F17-Oscillations%2FSet-F%2FOscillations.pdf
Thanks!
paper: https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FPhysics%2FA-level%2FTopic-Qs%2FCAIE%2F17-Oscillations%2FSet-F%2FOscillations.pdf
Reply 1
you can think of it this way - the x-t graph is a sine graph from equilibrium position, so v-t graph will be a cosine graph and a-t graph will be a -sine graph since v=dx/dt and a=dv/dt. for max v, t is 0, T or 2T (assuming period is 2T), when seen on the x-t graph, these values correspond to 0, meaning the piston at equilibrium position has maximum velocity.
for max a, u can also use the same method to visualise it. since a = -sin and x = sin, max a occurs when x is max (ignoring the -ve sign since it js tells the direction of travel).
alternatively, using the formula, v = +/- w (x0 2 - x2 )1/2 and a = -w2 x, where xo is maximum displacement and x is displacement at that time, max v occurs when x=0, when piston is at equilibrium. similarly, max a will occur when x is maximum, when piston is at the top or bottom
knowing the positions of the piston, you can work out the positions of the pivot when the piston has max v and a
for max a, u can also use the same method to visualise it. since a = -sin and x = sin, max a occurs when x is max (ignoring the -ve sign since it js tells the direction of travel).
alternatively, using the formula, v = +/- w (x0 2 - x2 )1/2 and a = -w2 x, where xo is maximum displacement and x is displacement at that time, max v occurs when x=0, when piston is at equilibrium. similarly, max a will occur when x is maximum, when piston is at the top or bottom
knowing the positions of the piston, you can work out the positions of the pivot when the piston has max v and a
Reply 2
13
Original post
by helloosyou can think of it this way - the x-t graph is a sine graph from equilibrium position, so v-t graph will be a cosine graph and a-t graph will be a -sine graph since v=dx/dt and a=dv/dt. for max v, t is 0, T or 2T (assuming period is 2T), when seen on the x-t graph, these values correspond to 0, meaning the piston at equilibrium position has maximum velocity.
for max a, u can also use the same method to visualise it. since a = -sin and x = sin, max a occurs when x is max (ignoring the -ve sign since it js tells the direction of travel).
alternatively, using the formula, v = +/- w (x0 2 - x2 )1/2 and a = -w2 x, where xo is maximum displacement and x is displacement at that time, max v occurs when x=0, when piston is at equilibrium. similarly, max a will occur when x is maximum, when piston is at the top or bottom
knowing the positions of the piston, you can work out the positions of the pivot when the piston has max v and a
for max a, u can also use the same method to visualise it. since a = -sin and x = sin, max a occurs when x is max (ignoring the -ve sign since it js tells the direction of travel).
alternatively, using the formula, v = +/- w (x0 2 - x2 )1/2 and a = -w2 x, where xo is maximum displacement and x is displacement at that time, max v occurs when x=0, when piston is at equilibrium. similarly, max a will occur when x is maximum, when piston is at the top or bottom
knowing the positions of the piston, you can work out the positions of the pivot when the piston has max v and a
Sorry, I just wanted to ask, I was taught that the x-t graph is a cosine graph because x=x0 cos wt? thanks!
Reply 3
21
Original post
by anonymous56754Sorry, I just wanted to ask, I was taught that the x-t graph is a cosine graph because x=x0 cos wt? thanks!
Id guess sine refers to sinusodal as the general solution for an SHM x-t curve is
x(t) = Bcos(wt)+Csin(wt) = A cos(wt + phi)
where A,B,C,phi are coefficients that represent the amplitude/initial position/velocity/phase offset. So its a sinusoidal curve, where time and position are scaled and time is translated to reflect the initial conditions. If X(0) = A, so the initial posiiton is the max amplitude / zero initial velocity, then the simple way to write shm x-t is
x(t) = A cos(wt)
But its just a sin(usoidal) curve.
(edited 11 months ago)
Reply 4
Original post
by anonymous56754Sorry, I just wanted to ask, I was taught that the x-t graph is a cosine graph because x=x0 cos wt? thanks!
not sure if it makes a difference if i took the international A lvls (sg-cambridge), but as far as i know and am taught, x-t graph can either be sin or cos, it ultimately depends on the original position of the object/piston in the oscillation (when t=0). if it was a pendulum displaced from the rest position to the side, x-t wld be a sine graph, but if it was displaced from the side, x-t wld be a cosine graph.
anyway, whether x-t is sine or cosine wldnt matter much since we're only finding the positions of v max and a max. but since you were taught x-t as cos graph, maybe u shld js ignore the first method haha
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