Math Problem Statement

P diganti dengan 5

Calculate various wave properties for the given sinusoidal wave equation with P = 5.

Solution

The given wave equation is:

y(x,t)=12Psin(4πxPπt+π)y(x, t) = \frac{1}{2} P \sin(4 \pi x - P \pi t + \pi)

where PP is defined as the sum of the third digit and the last digit of a given identifier (NIM), which you specified to be replaced by P=5P = 5.

This makes the wave equation:

y(x,t)=125sin(4πx5πt+π)=52sin(4πx5πt+π)y(x, t) = \frac{1}{2} \cdot 5 \sin(4 \pi x - 5 \pi t + \pi) = \frac{5}{2} \sin(4 \pi x - 5 \pi t + \pi)

Let's address each question:

(a) Fase Awal Gelombang

The initial phase of the wave is the constant term inside the sine function, which here is π\pi. Thus, the initial phase is π\pi radians.

(b) Fase pada Posisi x=1.5mx = 1.5 \, \text{m} dan Waktu t=0.25st = 0.25 \, \text{s}

To find the phase at x=1.5x = 1.5 m and t=0.25t = 0.25 s, we substitute these values into the argument of the sine function: Phase=4π(1.5)5π(0.25)+π\text{Phase} = 4 \pi (1.5) - 5 \pi (0.25) + \pi Calculate each term to get the result.

(c) Kecepatan Perambatan Gelombang

The wave velocity vv for a sinusoidal wave is given by: v=ωkv = \frac{\omega}{k} where ω\omega is the angular frequency and kk is the wave number. For the given equation:

  • k=4πk = 4\pi
  • ω=5π\omega = 5\pi

Thus, v=5π4π=54m/sv = \frac{5 \pi}{4 \pi} = \frac{5}{4} \, \text{m/s}

(d) Kecepatan Transversal Maksimum

The maximum transverse velocity vmaxv_{\text{max}} is given by: vmax=Aωv_{\text{max}} = A \omega where A=52A = \frac{5}{2} and ω=5π\omega = 5 \pi. So, vmax=525π=25π2m/sv_{\text{max}} = \frac{5}{2} \cdot 5 \pi = \frac{25 \pi}{2} \, \text{m/s}

(e) Percepatan Transversal Maksimum

The maximum transverse acceleration amaxa_{\text{max}} is: amax=Aω2a_{\text{max}} = A \omega^2 Calculating it with A=52A = \frac{5}{2} and ω=5π\omega = 5 \pi, amax=52(5π)2=5225π2=125π22m/s2a_{\text{max}} = \frac{5}{2} \cdot (5 \pi)^2 = \frac{5}{2} \cdot 25 \pi^2 = \frac{125 \pi^2}{2} \, \text{m/s}^2

(f) Kecepatan Transversal di x=2mx = 2 \, \text{m} dan t=1st = 1 \, \text{s}

To find the transverse velocity, differentiate y(x,t)y(x, t) with respect to tt, then substitute x=2x = 2 m and t=1t = 1 s.

(g) Percepatan Transversal di x=2mx = 2 \, \text{m} dan t=1st = 1 \, \text{s}

Differentiate the transverse velocity with respect to tt to find the acceleration, then substitute x=2x = 2 m and t=1t = 1 s.

(h) Gambarkan Gelombang

To plot the wave, one would graph y(x,t)y(x, t) versus xx or tt, showing a sinusoidal shape with parameters defined above.

Would you like more detailed calculations or further explanation for any part?


Related Questions

  1. How do you find the wave equation if PP changes?
  2. What is the significance of initial phase in wave motion?
  3. How does changing ω\omega or kk affect wave velocity?
  4. What is the difference between phase and phase velocity?
  5. How can transverse acceleration be practically observed in a wave?

Tip

To analyze waves more efficiently, remember that phase velocity vv can be derived from the ratio of angular frequency to wave number.

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Math Problem Analysis

Mathematical Concepts

Wave Motion
Trigonometry
Calculus
Sinusoidal Waves

Formulas

Wave Equation: y(x, t) = A sin(kx - ωt + φ)
Wave Velocity: v = ω / k
Maximum Transverse Velocity: v_max = Aω
Maximum Transverse Acceleration: a_max = Aω^2
Phase Calculation: ϕ = kx - ωt + φ

Theorems

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Suitable Grade Level

Undergraduate Physics