SHM amplitude and angular frequency from velocity equations

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Homework Statement
The motion of a body along a straight line is simple harmonic. When 60 mm
from the mid-point of the motion, the velocity of the body is 12 ms–1 and when
90 mm from the mid-point, its velocity is 4 ms-1. For the motion, calculate the:
(a) Amplitude
(b) Angular Frequency
(c) Frequency
(d) Periodic time
Relevant Equations
v^2 = w^2 (A^2-x ^2)

where:

v = velocity (m/s)
w= angular frequency (rad/s)
A = amplitude (m)
x = displacement from the midpoint (m)
At
Handwritten equation x1 = 60 mm = 0.06 m, written in italic x with subscript 1.
,
v1 = 12m/s
12^2 = w^2 (A^2-0.06^2)
144 = w^2 (A^2 - 0.0036)



At
Handwritten equation: x2 = 90 mm = 0.09 m
,
v2 = 4 m/s
4^2 = w^2(A^2 - 0.09^2)
16 = w^2 (A^2 - 0.0081)




(b) Angular frequency.

Subtract the second equation from the first:

Handwritten subtraction of two equations: 144 - 16 = ω²(0.0081 - 0.0036).


Handwritten equation: 128 = ω²(0.0045), a rearranged step from the energy-type SHM relations for angular frequency and amplitude.


Handwritten equation: omega squared equals 128 divided by 0.0045, equals 28444.44.


Handwritten equation: omega = 168.7 rad/s, with omega underlined.




(a) Amplitude

Substitute
1785436154830.webp
into one of the equations:

Handwritten equation: 16 = w^2(A^2 - 0.0081), with the value 28444.44 shown in place of w^2 in the printed equation.


Handwritten arithmetic: A squared minus 0.0081 equals 16 over 28444.44, which equals 0.0005625.


Handwritten equation A^2 = 0.0086625, the amplitude-squared result from solving the SHM velocity equations.


Handwritten calculation giving amplitude A equals the square root of 0.0086625, which equals 0.0931 m.


Handwritten working showing the amplitude result A = 0.0931 m = 93.1 mm, written as a single line of mathematics.




(c) Find the frequency

Equation f = ω/(2π) written in plain form, relating frequency f to angular frequency ω.


Handwritten equation f = 168.7 / (2π) = 26.85 Hz, computing a frequency from an angular frequency value of 168.7.


Handwritten or typeset underlined expression reading f = 26.85 Hz.


​
(d) Periodic time

Equation T = 1/f, with T on the left and 1 over f on the right, written as a fraction.


Handwritten calculation: T = 1/26.85 = 0.0372 s, shown as a fraction equal to 0.0372 seconds.


Handwritten equation T = 0.0372 s, with T as the period in seconds.



Do they look alright for the answers?
Does it matter that i have solved them out of order for the question or should that be ok for my tutor ?

Thanks for your help.
 
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  • Handwritten equation: 4^2 = ω^2(A^2 − 0.09^2)
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Your work looks correct, but you might be penalised for quoting too many significant figures in your answers. Given a value like "90mm", you should only assume it is accurate to the nearest mm.
If you use the numeric output of one step as an input to the next, though, you should use the unrounded value as the input. E.g. you could answer 170 rad/s to part b then use your 28444 (as its square) for the next step.

But as a matter of technique, I strongly advise working purely algebraically as far as possible, avoiding plugging in numbers until the end. It has many advantages. For example, it avoids the accumulation of errors that can arise from such chains of calculation. So in part a, you would start again with your base equation and manipulate it eliminate A to get an equation for omega.

I see no need to solve in the same order as the questions are posed.
 
You can write your relevant equation twice, for each of the given pairs ##(x_1,v_1)## and ##(x_2,v_2)## to get a system of 2 equations and 2 unknowns, ##A^2## and ##\omega^2##.

Solve the system the usual way to find the unknowns, the order doesn't matter. I checked your value for ##\omega## and it is correct. I did not check the other values.

I second @haruspex's recommendation to avoid plugging numbers until the very end. Also, if you know how to use a spreadsheet, use it instead of a calculator. It makes troubleshooting your work easier and you can also set it to give the answer in the correct number of significant figures.
 
One of the things you'll need to learn in the real world, after you leave school, is that there usually is no answer key, and there often isn't anyone to ask. Either they don't know how to solve it, or they are too busy doing their own work. They didn't hire you to ask them for the answers, they want you to find the answers. So, being able to look back at the problem and checking your own work is a crucial skill.

This problem is a great example. You aren't really done when you get an answer. You're done when you've verified it for yourself. Asking about the underlying physics, or the correct equations that describe it, makes sense to me. Asking if the numbers are correct seems lazy. If you can find the answer, you can check the answer yourself.