SHM problem

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!

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!

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 !

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?

I think this must be undergrad!!

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 :/

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 !

Your answer and the book's answer are not incompatible.
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.
Your answer and the book's add to give 2pi rads (give or take rounding errors) or 360 degs.
It's the difference between these two possible ways of thinking of the motion regarding the initial displacement.

Phase angle phi in blue (arrow)

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.