Physics Question

Acceleration at the equilibrium position is easy - SHM occurs when there is a restoring force pointing towards the equilibrium position AND the magnitude of the restoring force is proportional to the distance from the equilibrium position.
when the mass is at the equilibrium position the distance from the equilibrium position is zero... and F=ma tells us that the acceleration here must also be zero.

for the other part you will need to select an appropriate formula from the data sheet and convert the information given by the question into the appropriate SI units

is there a circular motion formula that can help you find the amount of centripetal force for the car?

is there a circular motion formula that can help you find the amount of centripetal force for the car?

??helloo

So what's the angle at which the vertical component of the normal force is equal in size to the weight force and the horizontal component is equal in size to the centripetal force?

So what's the angle at which the vertical component of the normal force is equal in size to the weight force and the horizontal component is equal in size to the centripetal force?

Have a look at this - you should have done a sketch that looks something like diagram 3 to help you.

https://study.com/academy/lesson/circular-motion-around-a-banked-circular-track.html

So am I using the formula Nsintheta=mv^2/R

That's just a stepping stone... it's got N and sin θ in it - and you don't know N yet but you want θ
you can keep going like it shows till you get
tan θ = v²/(gR)

or you can just look at the sketch diagram you've made which looks something like this


and realise that you can get tan θ already because you already know Ny... because it's the size of the weight force on the car
and you already know Nx because it's the centripetal force on the car.

so you can just read off the sketch that tan θ = centripetal force / weight force
and you already know numerical values for both those.

That's why drawing sketches is a really useful thing to remember to do.

f


for v do i use angular speed or normal speed?

it's the linear speed in m/s, careful about using normal to mean usual cos normal also means perpendicular
in the circular motion types of formulae, angular velocity is always ω (lower case omega)

I have tried working it out I get very low answer between 0-1

OK
circumference of the rack is 2πr = 1570m
speed of the vehicle is d/t = 52.3 m/s

tan θ = v²/(gr)
=2740/(250*9.81)
=1.12

1.12 is the tangent of the angle we're looking for so...
tan^-1 1.12 = 48.2°

Thanks I was using different formulas to work it out and I didn’t take into account the circumference

tan^-1 does not equal 48.2 it is 0.84