The equation y(x,t)=Acos2πf(xv−t) may be written as y(x,t)=Acos[2πλ(x−vt)].

**Answer 1**

Part A

y(x,t) is the particle’s “transverse” location. (perpendicular to the wave’s direction, as opposed to x- the particle’s longitudinal location) To get transverse velocity, just take the partial derivative of y(x,t) with respect to time:

d/dt {y(x,t)} = d/dt {A cos[2 π / λ (x – v t)]}

d/dt {y(x,t)} = – A sin[2 π / λ (x – v t)] d/dt {2 π / λ (x – v t)}

d/dt {y(x,t)} = – A sin[2 π / λ (x – v t)] 2 π / λ (-v)

d/dt {y(x,t)} = A 2 π / λ v sin[2 π / λ (x – v t)]

Part B

Now you need to find when the second derivative with respect to time (the transverse acceleration) is zero. That’s when the velocity is at it’s maximum speed:

d²/dt² {y(x,t)} = d/dt {A 2 π / λ v sin[2 π / λ (x – v t)]}

d²/dt² {y(x,t)} = A 2 π / λ v d/dt {sin[2 π / λ (x – v t)]}

d²/dt² {y(x,t)} = A 2 π / λ v cos[2 π / λ (x – v t)] d/dt {2 π / λ (x – v t)}

d²/dt² {y(x,t)} = A 2 π / λ v cos[2 π / λ (x – v t)] 2 π / λ (-v)

d²/dt² {y(x,t)} = – A 4 π² / λ² v² cos[2 π / λ (x – v t)]

0 = – A 4 π² / λ² v² cos[2 π / λ (x – v t)]

0 = cos[2 π / λ (x – v t)] . . . . .

cos(Θ) = 0 at Θ = 0 and π. We can use either, because the speed is max at either:

2 π / λ (x – v t) = 0

x – v t = 0

t = x / v

Now insert this time into the equation for the velocity in from Part A:

d/dt {y(x,t)} = A 2 π / λ v sin[2 π / λ (x – v t)]

d/dt {y(x,t)} = A 2 π / λ v sin[2 π / λ (x – v (x / v))]

d/dt {y(x,t)} = A 2 π / λ v sin[2 π / λ (x – x)]

d/dt {y(x,t)} = A 2 π / λ v sin[2 π / λ (0)]

d/dt {y(x,t)} = A 2 π / λ v sin[0]

sin(Θ) = 0 at Θ = π/2 and 3π/2. We can use either, because the speed is max at either:

d/dt {y(x,t)} = A 2 π / λ v (π/2)

d/dt {y(x,t)} = A π² / λ v

Of course, now that I’ve worked all of this out I realize there’s a pretty obvious shortcut, if you’re allowed to take it. For a wave, the transverse velocity is maximum when acceleration AND displacement are zero. This is true for pendulums or any other kind of simple harmonic motion. Rather than taking the second derivative, you could just solve the original y(x,t) equation for t when y(x,t) = 0:

y(x,t) = A cos[2 π / λ (x – v t)]

0 = A cos[2 π / λ (x – v t)]

0 = cos[2 π / λ (x – v t)]

2 π / λ (x – v t) = 0

x – v t = 0

t = x / v

Duh! From there, the rest of the solution for Part B is the same.

**Answer 2**

For anyone else still trying to find this question….

For part B, take the maximum value of sin (which is 1) and multiply it by the constant you find in part A, which is the derivative (d/dt) for the second given equation.

**Answer 3**

cos(theta)=0 at theta=pi/2 or 3pi/2, not theta=0 or pi

**Answer 4**

Part A:

vy = (2πv/λ)Asin(2πλ(x−vt))

Part B:

vmax =(2πv/λ)A

Source(s): My head

**Answer 5**

i don’t understand about it.