SHM: Choose Sin or Cos to Solve?

[Isweer]

I wanted to ask if weather I choose sin or cos makes difference in solving problems.
In this question

A block of mass m=4 kg on a frictionless horizontal table is attached to one end of a spring of force constant k=400 N/m and undergoes simple harmonic oscillations about its equilibrium position (X=0) with amplitude A=6 cm. If the block is at x=6 cm at time t=0, then which of the following equations (x in cm and t in seconds) gives the block's position as a function of time?

a) x=6sin(10t + ∏/2)
b) x=6sin(10t - ∏/2)
c)...
d)...
e)...

the answer is (a) but, what I am asking is that I got cos instead of sin ( from the general form X=Acos(ωt + ∅ )

 

[Integral]

If you could show us some of your work, we could help you.

 

[Isweer]

When I started solving this problem I immediately wrote the general form I learned.
X=A*cos(wt + ∅ ).

Then I started breaking the problem into pieces:

A=6,
w= (k/m)^0.5 = (400/4)^0.5 = 10,
and ∅= sin(inverse) (x at t=0)/A)= sin(inverse) (6/6)= 90 degrees which is ∏/2

by substituting I got:

X=6cos(10t + ∏/2)

so does that mean that using sin or cos doesn't matter? or I have a mistake?

 

[Curious3141]

When I started solving this problem I immediately wrote the general form I learned.
X=A*cos(wt + ∅ ).

Then I started breaking the problem into pieces:

A=6,
w= (k/m)^0.5 = (400/4)^0.5 = 10,
and ∅= sin(inverse) (x at t=0)/A)= sin(inverse) (6/6)= 90 degrees which is ∏/2

by substituting I got:

X=6cos(10t + ∏/2)

so does that mean that using sin or cos doesn't matter? or I have a mistake?

You've made a mistake. Why did you do this bit?

and ∅= sin(inverse) (x at t=0)/A)= sin(inverse) (6/6)= 90 degrees which is ∏/2

Why sine?

When you let t = 0 in $6\cos({\omega}t + \varphi)$, you get $6\cos\varphi = 6$, giving $\varphi=0$. So the equation should simply be $6\cos{\omega}t$.

This is not one of the answers. However, you should be able to see (from basic trig) that $\sin(\theta + \frac{\pi}{2}) = \cos\theta$. So you can match it to the first answer.

It doesn't matter if you use sine or cosine, both are equally valid. They may look different, but if you do the basic math, they're actually identical.

And here, you don't actually have to derive very much. It's easier to just go through the multiple choices to see which fits, i.e. which gives a displacement of 6 when t = 0.

 

[Isweer]

Thanks for help