# SHM problemWatch

#1
Hello, I wish you happiness and well-being.

Bit stuck on this Physics problem, can anyone help? I attached the pic to this post. I have gotten Amplitude and angular frequency correctly but keep getting a supposedly wrong answer for phi (I'm getting 0.504 rads but the answer is 2.7 rads) can someone verify which is the correct answer and give some hints as to how to get 2.7 if it is the correct answer?

Thanks!

0
4 years ago
#2
Hi! I gave part a a go, I am not sure this is right so don't quote me!! Total energy can be equated to energy when v is max, I see v is 0.5ms-1, so 0.5 X Mass X v^2 = 0.5 X 075 X 0.5^2 = 0.09375 J

Can someone else chime in here and say if this is right or not !

(Original post by GuanyinBuddha)
Hello, I wish you happiness and well-being.

Bit stuck on this Physics problem, can anyone help? I attached the pic to this post. I have gotten Amplitude and angular frequency correctly but keep getting a supposedly wrong answer for phi (I'm getting 0.504 rads but the answer is 2.7 rads) can someone verify which is the correct answer and give some hints as to how to get 2.7 if it is the correct answer?

Thanks!

0
4 years ago
#3
(Original post by GuanyinBuddha)
Hello, I wish you happiness and well-being.

Bit stuck on this Physics problem, can anyone help? I attached the pic to this post. I have gotten Amplitude and angular frequency correctly but keep getting a supposedly wrong answer for phi (I'm getting 0.504 rads but the answer is 2.7 rads) can someone verify which is the correct answer and give some hints as to how to get 2.7 if it is the correct answer?

Thanks!

I am getting 2.6 rads tan of phi should equal minus the constant of the sine expression/ over the constant of the cos expression due to the expansion of rcos(x+a).What working out have you done?
0
4 years ago
#4
I think this must be undergrad!!
0
#5
(Original post by Hexaneandheels)
Hi! I gave part a a go, I am not sure this is right so don't quote me!! Total energy can be equated to energy when v is max, I see v is 0.5ms-1, so 0.5 X Mass X v^2 = 0.5 X 075 X 0.5^2 = 0.09375 J

Can someone else chime in here and say if this is right or not !
The total energy is 0.41375J
1
#6
(Original post by Dalek1099)
I am getting 2.6 rads tan of phi should equal minus the constant of the sine expression/ over the constant of the cos expression due to the expansion of rcos(x+a).What working out have you done?

Darn...

Well they give you the x the displacement, what I did was differentiate Acos(wt+phi) to get -Awsin(wt+phi) = V correct?

At t = 0, V = 0.5ms-1 so => sin(phi) =-V/(Aw) and I got the wrong answer :/
0
#7
(Original post by Hexaneandheels)
I think this must be undergrad!!
I marked it as undergrad, but I think a sixth former/college student could also answer it if they understood SHM
0
4 years ago
#8
(Original post by GuanyinBuddha)
Darn...

Well they give you the x the displacement, what I did was differentiate Acos(wt+phi) to get -Awsin(wt+phi) = V correct?

At t = 0, V = 0.5ms-1 so => sin(phi) =-V/(Aw) and I got the wrong answer :/
start off with setting the general solution as Acoswt +Bsinwt and Acoswt=0.04 and note A is not the amplitude here then differentiate Bwcoswt=0.50 at t=0
and tanphi=-B/A from equating terms with expansion of Rcos(x+phi) adjust using period of tan as 180 and you should get 2.6 I don't know whether the book has rounded early or something.
1
4 years ago
#9
(Original post by GuanyinBuddha)
***
It just depends whether you consider the 4cm initial displacement to be positive or negative.
The book's answer has taken it to be -4cm.
It's the difference between these two possible ways of thinking of the motion regarding the initial displacement.

Phase angle phi in blue (arrow)
4 years ago
#10
(Original post by Hexaneandheels)
Hi! I gave part a a go, I am not sure this is right so don't quote me!! Total energy can be equated to energy when v is max, I see v is 0.5ms-1, so 0.5 X Mass X v^2 = 0.5 X 075 X 0.5^2 = 0.09375 J

Can someone else chime in here and say if this is right or not !
Energy = Kinetic Energy + Potential Energy

Kinetic Energy = 0.5mv^2 = 0.5*0.75*0.5^2 = 0.09375

Potential Energy of a spring = 0.5kx^2, where k is spring constant and x is displacement.

That is,

Potential Energy = 0.5*400*0.04^2 (remember to convert cm to m as I have done here)

Potential Energy = 0.32

Therefore,

Total Energy = 0.09375 + 0.32 = 0.41375 J
0
#11
(Original post by Stonebridge)
It just depends whether you consider the 4cm initial displacement to be positive or negative.
The book's answer has taken it to be -4cm.
It's the difference between these two possible ways of thinking of the motion regarding the initial displacement.

Phase angle phi in blue (arrow)
Thank you very much, that was an extremely helpful answer especially the diagrams
0
#12
(Original post by Dalek1099)
start off with setting the general solution as Acoswt +Bsinwt and Acoswt=0.04 and note A is not the amplitude here then differentiate Bwcoswt=0.50 at t=0
and tanphi=-B/A from equating terms with expansion of Rcos(x+phi) adjust using period of tan as 180 and you should get 2.6 I don't know whether the book has rounded early or something.

Thanks mate, forgot I could also solve it like that
0
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