To see how two traveling waves of the same frequency create a standing wave.
Consider a traveling wave described by the formula
y1(x,t)=Asin(kx−ωt).
This function might represent the lateral displacement of a string, a local electric field, the position of the surface of a body of water, or any of a number of other physical manifestations of waves.
At certain times, the string will be perfectly straight. Find the first time t1>0 when this is true.
Express t1 in terms of ω, k, and necessary constants.
The answer is t1=pi/2ω
can you please explain how this came to be? thanks
The equation of standing wave is y(x,t)=(2Asinkx)cosωt.
The string can be straight only when cosωt =0, for then y(x,t)=0 also (for all ‘x’)
If cosωt₁ =0, ωt₁=π/2, => t₁=π/2ω
Well, let's dive into the waves and see what's going on!
For a traveling wave, the highest (or lowest) points of the wave occur at specific values of the phase. In this case, when the string is straight, it means the wave is at its peak or trough at that particular point.
To find the first time when the string is perfectly straight, we need to find the time when the phase of the wave is an integer multiple of π. In other words, we want the expression inside the sine function to be an integer multiple of π.
So, we set (kx - ωt) equal to an integer multiple of π.
kx - ωt = nπ,
where n is an integer.
Now, to find the first time t1, we want to find the smallest positive value of t that satisfies this equation.
Rearranging the equation, we have:
t = (kx - nπ)/ω.
Since we want to find the first time when the string is straight, we set n = 0 to get:
t1 = (kx - 0)/ω,
t1 = kx/ω.
Now, recall that the wave is traveling, so its speed is given by v = ω/k. We can rewrite the equation as:
t1 = (kx)/(ω/k),
t1 = (kx)/(v),
t1 = x/v.
This means that the first time t1 when the string is perfectly straight is just the time it takes for the wave to travel a distance x, which is given by x/v.
Now, since we are dealing with sine waves, we know that when t = π/2ω, we have completed a quarter of a period. So, substituting t1 = π/2ω, we have:
π/2ω = x/v.
Rearranging the equation, we get:
x = (π/2ω)v.
And since v = ω/k, we can substitute it back in:
x = (π/2ω)(ω/k),
x = πk/2.
Therefore, the first time t1 when the string is perfectly straight is:
t1 = x/v,
t1 = (πk/2)/(ω/k),
t1 = π/2ω.
So, the answer is t1 = π/2ω.
I hope this explanation didn't tie you up in knots!
To find the first time when the string is perfectly straight, we need to find the time at which the displacement function y1(x, t) becomes zero.
We start by setting y1(x, t) = 0:
Asin(kx - ωt) = 0
Since the sine function is zero when its argument is an integer multiple of π, we have:
kx - ωt = nπ
Where n is an integer.
Now, we are interested in finding the first time t1 when the string is straight. This means that x can take any value and kx will be constant. Let's denote kx as constant K:
kx = K
Substituting this into the equation above:
K - ωt = nπ
Rearranging the equation:
ωt = K - nπ
To find the first time t1 when the string is straight, we want to find the smallest positive value of t that satisfies this equation. This occurs when the right-hand side is equal to zero:
K - nπ = 0
Solving for K:
K = nπ
Substituting K back into the equation for ωt:
ωt = nπ
Solving for t:
t = nπ/ω
Since we are looking for the smallest positive value of t, we take n = 1:
t1 = (π/ω)
Therefore, the first time t1 when the string is perfectly straight is t1 = (π/ω), which matches the given answer.
To find the first time t1 when the string is perfectly straight, we need to determine the condition when the y1(x, t) function equals zero. In other words, we need to solve for x and t in the equation y1(x, t) = Asin(kx − ωt) = 0.
For a sine function to be equal to zero, the argument inside the sine function (kx − ωt) must be equal to an integer multiple of π. This means:
kx − ωt = nπ,
where n is an integer.
To find the first time t1 when the string is perfectly straight, we want to find the smallest positive value of t that satisfies the above equation. Let's solve it for t:
ωt = kx + nπ,
t = (kx + nπ) / ω.
Since we want to find the first time t1 when the string is straight, we can set x = 0 to simplify the expression:
t1 = (k(0) + nπ) / ω,
t1 = nπ / ω.
To find the smallest positive value of t1, we set n = 1:
t1 = 1π / ω,
t1 = π / ω.
Therefore, the answer is t1 = π / ω.