for part d, why can't we use: V^2 = (w^2)(a^2 = x^2)
its fastest at the equilibrium point O, but at that point the extension is still 0.3.... ?
its fastest at the equilibrium point O, but at that point the extension is still 0.3.... ?
also, why is the amplitude for this spring 0.4 (which is the extension) and not the distance between the centre/equilibrium position and the particle (0.5), when the amplitude for the string above is the distance beyond the position of equilibrium/centre (0.2)?
@Notnek @mqb2766 @old_engineer @ghostwalker @DFranklin @RDKGames
is this something one of you guys can help me with please?
is this something one of you guys can help me with please?
Probably easier to go through this one by one, can you post your working and repost your first question pls?
The speed should just be
amplitude*frequency
The speed should just be
amplitude*frequency
Original post by Maths&physics
@Notnek @mqb2766 @old_engineer @ghostwalker @DFranklin @RDKGames
is this something one of you guys can help me with please?
is this something one of you guys can help me with please?
(edited 4 years ago)
Original post by mqb2766
Probably easier to go through this one by one, can you post your working and repost your first question pls?
The speed should just be
amplitude*frequency
The speed should just be
amplitude*frequency
for the first question, when calculating the fastest speed of SHM, this is when it passes through the equilibrium point (when extension is 0). however, the extension is still 0.3 at the equilibrium point.
working out: V^2 = (80/7)[(0.2)^2 - (0.3)^2]
and for the second question, the spring has been pushed/pulled beyond the equilibrium point. therefore, I thought the amplitude was 0.5 and not 0.4 (which is the extension). there was no working out for this question.
At the equilibrium point the extension is zero ant the velocity is maximum.
x = Acos (wt)
dx/dt = -Awsin(wt)
When cos() is zero, sin is max and speed is equal to Aw.
Is this ok.
x = Acos (wt)
dx/dt = -Awsin(wt)
When cos() is zero, sin is max and speed is equal to Aw.
Is this ok.
Original post by Maths&physics
for the first question, when calculating the fastest speed of SHM, this is when it passes through the equilibrium point (when extension is 0). however, the extension is still 0.3 at the equilibrium point.
working out: V^2 = (80/7)[(0.2)^2 - (0.3)^2]
and for the second question, the spring has been pushed/pulled beyond the equilibrium point. therefore, I thought the amplitude was 0.5 and not 0.4 (which is the extension). there was no working out for this question.
working out: V^2 = (80/7)[(0.2)^2 - (0.3)^2]
and for the second question, the spring has been pushed/pulled beyond the equilibrium point. therefore, I thought the amplitude was 0.5 and not 0.4 (which is the extension). there was no working out for this question.
Original post by mqb2766
At the equilibrium point the extension is zero ant the velocity is maximum.
x = Acos (wt)
dx/dt = -Awsin(wt)
When cos() is zero, sin is max and speed is equal to Aw.
Is this ok.
x = Acos (wt)
dx/dt = -Awsin(wt)
When cos() is zero, sin is max and speed is equal to Aw.
Is this ok.
ok, now that makes sense. it's tension that doesn't = 0 in both of these questions. you're a legend! thank you so much!
No problem, I've not even looked at the second question but it sounds like I don't need to so ...
Original post by Maths&physics
ok, now that makes sense. it's tension that doesn't = 0 in both of these questions. you're a legend! thank you so much!
Original post by mqb2766
No problem, I've not even looked at the second question but it sounds like I don't need to so ...
oh yeah. could you please because I think the textbook made a mistake.
Shouldn't have asked :-). Can you upload your working pls?
Original post by Maths&physics
oh yeah. could you please because I think the textbook made a mistake.
Could you upload the answer in the book as well?
Original post by Maths&physics
oh yeah. could you please because I think the textbook made a mistake.
Original post by mqb2766
Shouldn't have asked :-). Can you upload your working pls?
I didn't have any working out for that question. here is the question (part b asking for amp) and here is their working. I thought the amp was 0.5 because that's how far its been stretched/pulled beyond the equilibrium point.
(edited 4 years ago)
Fully agree. The amplitude (coefficients of the sin(wt) and cos(wt) terms) is determined by the initial conditions. This is the usual case where the velocity is 0 at t=0 and the displacement is 0.5 (given), so
x = 0.5cos(wt)
could be +/- depending on the definition of x, but would not affect the amplitude which is 0.5. The 0.4 in the model solution seems to appear out of nowhere and is wrong.
x = 0.5cos(wt)
could be +/- depending on the definition of x, but would not affect the amplitude which is 0.5. The 0.4 in the model solution seems to appear out of nowhere and is wrong.
Original post by Maths&physics
I didn't have any working for that question. here is the question (part b asking for amp) and here is their working. I thought the amp was 0.5 m because that's how far its been stretched/pulled beyond the equilibrium point.
Original post by mqb2766
Fully agree. The amplitude (coefficients of the sin(wt) and cos(wt) terms) is determined by the initial conditions. This is the usual case where the velocity is 0 at t=0 and the displacement is 0.5 (given), so
x = 0.5cos(wt)
could be +/- depending on the definition of x, but would not affect the amplitude which is 0.5.
x = 0.5cos(wt)
could be +/- depending on the definition of x, but would not affect the amplitude which is 0.5.
sadly, there have been so many mistakes in the textbook.
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