Oscillatory motion equation in sine function

[grangr]

Homework Statement

The equation y = A sin(kx - wt + pi/2) is the same as

a. y = -A sin(kx - wt + pi/2)
b. y = A cos(kx - wt)
c. y = -A cos(kx - wt)
d. y = -A sin(kx - wt - pi/2)
e. y = A sin(kx - wt + (3pi)/2)​

Homework Equations

• y = A sin[(2pi)/lamda * x - (2pi)/period * t + (phase constant)]
• sin(x) = cos(pi/2 - x)

The Attempt at a Solution

My attempted answer was d., while the correct answer given was b.

I do not understand why d. is wrong, as after a left shift of pi (from the '-pi/2' in the phase constant) in the sine function, when the sign is inverted (given the '-' before A), the resulting y should be the same. No?

I do not get why b. is the correct answer, either. Even with the fact that sin(x) = cos(pi/2 - x), wouldn't you get y = A cos(-kx + wt) then?Thanks in advance for your help!

 

[Nguyen Son]

Here's an answer

 

[rude man]

Here's an answer

OK, perhaps the long way around.
Realize sin(a+ pi/2) = cos(a)
a = kx-wt

 

[Nguyen Son]

I did that because he/she knows that sin(x) = cos(pi/2 - x), like he/she said above, but he/she didn't know how to solve this problem with sin(x) = cos(pi/2 - x), so I did in that way :D

 

[rude man]

I did that because he/she knows that sin(x) = cos(pi/2 - x), like he/she said above, but he/she didn't know how to solve this problem with sin(x) = cos(pi/2 - x), so I did in that way :D

Sorry, I thought your post was the OP's answer. But you should not post complete answers, just hints.

 

[PeroK]

Homework Statement

The equation y = A sin(kx - wt + pi/2) is the same as

a. y = -A sin(kx - wt + pi/2)
b. y = A cos(kx - wt)
c. y = -A cos(kx - wt)
d. y = -A sin(kx - wt - pi/2)
e. y = A sin(kx - wt + (3pi)/2)​

Homework Equations

• y = A sin[(2pi)/lamda * x - (2pi)/period * t + (phase constant)]
• sin(x) = cos(pi/2 - x)

The Attempt at a Solution

My attempted answer was d., while the correct answer given was b.

I do not understand why d. is wrong, as after a left shift of pi (from the '-pi/2' in the phase constant) in the sine function, when the sign is inverted (given the '-' before A), the resulting y should be the same. No?

I do not get why b. is the correct answer, either. Even with the fact that sin(x) = cos(pi/2 - x), wouldn't you get y = A cos(-kx + wt) then?Thanks in advance for your help!

$\cos(\theta) = \sin(\theta + \frac{\pi}{2}) = - \sin(\theta - \frac{\pi}{2})$

So, both b) and d) are correct.

 

[grangr]

Thank you all, Nguyen Son, rude man, and Perok for your explanation. It is much clearer now. (And it's good to know that both b) and d) are correct! )

Indeed, I totally did not recall that cos(x) = cos(-x), or that cos(θ)=sin[θ+(π/2)]. As it turns out, it wasn't a physics problem to me, but more of a math problem...