# A question in SHM ?

1. Jan 28, 2012

### Isweer

I wanted to ask if weather I choose sin or cos makes difference in solving problems.
In this question
the answer is (a) but, what I am asking is that I got cos instead of sin ( from the general form X=Acos(ωt + ∅ )

2. Jan 28, 2012

### Integral

Staff Emeritus

3. Jan 28, 2012

### 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?

4. Jan 28, 2012

### Curious3141

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

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.

5. Jan 29, 2012

### Isweer

Thanks for help