Answer :
Answer:
d/dt {y(x,t)} = A 2 π / λ v sin[2 π / λ (x - v t)]
see below for further answers and explanation
Explanation:
part 1 of the equation should be written as thus
y(x,t)=Acos2πf((x/v)-t)
Take partial derivative
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)]
2. The second equation can be rearranged as thus:
y(x,t)=Acos[2π/λ(x-vt)]
We take second derivative with respect to time
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)]
when speed is at maximum, theta=0 and pi=0, either of the two
2 π / λ (x - v t) = 0
x - v t = 0
t = x / v
Going to part 1 , insert the value of t into the equation, we have
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
The transverse velocity of the particle is: [tex]\mathbf{v_y =\dfrac{2 \pi v}{\lambda } A sin \Big[ \dfrac{2 \pi}{\lambda } \Big(x -vt \Big) \Big]}[/tex]
The maximum speed of a particle of the string is [tex]\mathbf{v_{y \ max} = \dfrac{2 \pi v A}{\lambda }}[/tex]
To solve this question, we need to understand the concept of a transverse wave in a string and the transverse velocity along with a wave.
What is a transverse wave?
A transverse wave is a wave whose own oscillations are at a right angle to the path in which the wave propagation travels.
The transverse wave can be expressed by using the formula:
[tex]\mathbf{y(x,t) = Acos \Big[ \dfrac{2 \pi}{\lambda} (x - v t) \Big]}[/tex]
Here;
- The wave is = y(x,t)
- The amplitude = A
- Wavelength = λ
- Frequency of wave = v
- The position(distance) of the wave = x
- Period of the wave = t
Similarly, the velocity of the partcle in which the wave travels can be expressed by using the relation:
- [tex]\mathbf{v_y = v(x,t)}[/tex]
Taking the differential of vy with respect to t, we have:
- [tex]\mathbf{v_y =\dfrac{dy(x,t)}{dt}}[/tex]
Also, the maximum velocity [tex]\mathbf{v_{y \ max}}[/tex] is computed as:
- [tex]\mathbf{v_{y \ max} = \dfrac{2 \pi AV}{\lambda}}[/tex]
(a)
From the equation of transverse wave, let replace the equation for y(x,t) and take the differentiation in order to determine the transverse velocity of the particle in the string on which the wave travels.
i.e.
[tex]\mathbf{v_y = \dfrac{d}{dt} \Big( A cos \Big[\dfrac{2 \pi}{\lambda }(x -vt) \Big] \Big)}[/tex]
By taking the differential, we can conclude that the transverse velocity of the particle is: [tex]\mathbf{v_y =\dfrac{2 \pi v}{\lambda } A sin \Big[ \dfrac{2 \pi}{\lambda } \Big(x -vt \Big) \Big]}[/tex]
(b)
To determine the maximum speed, we need to know that the speed of a particle includes the use of a sine form whose bound is from -1 to +1.
From the relation of the velocity of the particle:
- Using the +1 since form, the maximum value of the speed of a given frequency and wavelength relates to the term [tex]\mathbf{sin \Big[ \dfrac{2 \pi }{\lambda}(x - vt)\Big]}[/tex]
Hence,
[tex]\mathbf{v_{y \ max} = \dfrac{2 \pi v}{\lambda }\times A(1)}[/tex]
[tex]\mathbf{v_{y \ max} = \dfrac{2 \pi v A}{\lambda }}[/tex]
Learn more about transverse waves here:
https://brainly.com/question/11690308