Solve Sine Wave Function Homework

[Nanu Nana]

Homework Statement

A sine function is given.And the question is At which time does the deflection (y) reaches a maximum?
y= 3sin (800πxt+π/3)

Homework Equations

y= A sin (wt+Φ)
A=amplitude
w= angular frequency
Φ= initial phase

3. The Attempt at a Solution
I already have solution. But I don't understand it .
800πt+π/3 =π/2 +kx2xπ
800πt=π/6 + k x2xπ
t=1/4800 + k x 1/400
What does that k stands for ? and why is it kx2xπ and π/2 why not π/6

 

[PeroK]

Homework Statement

A sine function is given.And the question is At which time does the deflection (y) reaches a maximum?
y= 3sin (800πxt+π/3)[/B]

Homework Equations

y= A sin (wt+Φ)
A=amplitude
w= angular frequency
Φ= initial phase
3. The Attempt at a Solution
I already have solution. But I don't understand it .
800πt+π/3 =π/2 +kx2xπ
800πt=π/6 + k x2xπ
t=1/4800 + k x 1/400[/B]
What does that k stands for ? and why is it kx2xπ and π/2 why not π/6

When is the sine function at a maximum? If you have $y = 3sin(x)$ for what value(s) of $x$ is $y$ a maximum?

 

[Nanu Nana]

3 ?

 

[PeroK]

3 ?

3 is the maximum value of that function, but for which values of $x$ does it have that value?

 

[drvrm]

Homework Statement

A sine function is given.And the question is At which time does the deflection (y) reaches a maximum?
y= 3sin (800πxt+π/3)[/B]

Homework Equations

y= A sin (wt+Φ)
A=amplitude
w= angular frequency
Φ= initial phase
3. The Attempt at a Solution
I already have solution. But I don't understand it .
800πt+π/3 =π/2 +kx2xπ
800πt=π/6 + k x2xπ
t=1/4800 + k x 1/400[/B]
What does that k stands for ? and why is it kx2xπ and π/2 why not π/6

your y is function of t and you are writing condition for it -
you know that a sine function oscillates between zero and 1 - so you are putting the condition as first maxima will be at pi/2, but for a wave there will be second, third maxima of y...and so on ,that is written as additional term where k=0,1,2,... can be substituted and again one can get maxima- as after 2.pi addition one again reaches the max. y value... one can draw a diagram of y- phase curve and see how it repeats

so you get first maxima at t=1/4800 and second at t=1/4800 +1/400 (k=1) , and so on...

 

[Nanu Nana]

Thank you .

 

[PeroK]

your y is function of t and you are writing condition for it -
you know that a sine function oscillates between zero and 1 - so you are putting the condition as first maxima will be at pi/2, but for a wave there will be second, third maxima of y...and so on ,that is written as additional term where k=0,1,2,... can be substituted and again one can get maxima- as after 2.pi addition one again reaches the max. y value... one can draw a diagram of y- phase curve and see how it repeats

so you get first maxima at t=1/4800 and second at t=1/4800 +1/400 (k=1) , and so on...

That's the answer given on plate. However, you should note that:

1) Sine oscillates between -1 and 1 (not 0 and 1).

2) Sine has maxima for negative integer $k$ in addition to posiive integers.