Math Problem Statement

P diganti angka 3

Soal Fisika Gelombang: Gelombang sinusoidal merambat pada tali dengan persamaan y(x, t) = (1/2)P sin(4πx − Pπt + π), di mana P adalah digit ke-3 + digit terakhir NIM. Hitung fase awal, fase pada posisi tertentu, kecepatan perambatan, kecepatan transversal maksimum, percepatan transversal maksimum, dan lainnya dengan nilai P=3.

Solution

The problem involves a sinusoidal wave traveling on a string with the wave function:

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

where PP is a constant derived from the third digit and last digit of the student's NIM (student identification number). You specified that P=3P = 3.

So, we substitute P=3P = 3 into the equation, giving us:

y(x,t)=123sin(4πx3πt+π)=32sin(4πx3πt+π)y(x, t) = \frac{1}{2} \cdot 3 \sin(4\pi x - 3\pi t + \pi) = \frac{3}{2} \sin(4\pi x - 3\pi t + \pi)

Let's go through each part of the problem:

  1. Fase Awal Gelombang (Initial Phase of the Wave):

    • The initial phase, or the phase constant, is given by the term inside the sine function when x=0x = 0 and t=0t = 0. In this case, it’s π\pi.
  2. Fase pada Posisi x=1.5x = 1.5 meter dan Waktu t=0.25t = 0.25 detik:

    • Substitute x=1.5x = 1.5 and t=0.25t = 0.25 into the phase 4πx3πt+π4\pi x - 3\pi t + \pi: Phase=4π(1.5)3π(0.25)+π\text{Phase} = 4\pi (1.5) - 3\pi (0.25) + \pi
  3. Kecepatan Perambatan Gelombang (Wave Propagation Speed):

    • The wave speed vv can be found from the relationship v=ωkv = \frac{\omega}{k}, where k=4πk = 4\pi and ω=3π\omega = 3\pi.
  4. Kecepatan Transversal Maksimum (Maximum Transverse Velocity):

    • The maximum transverse velocity is given by vmax=Aωv_{\text{max}} = A\omega, where A=32A = \frac{3}{2} and ω=3π\omega = 3\pi.
  5. Percepatan Transversal Maksimum (Maximum Transverse Acceleration):

    • The maximum transverse acceleration is amax=Aω2a_{\text{max}} = A\omega^2.
  6. Kecepatan Transversal di x=2x = 2 meter dan t=1t = 1 detik:

    • Differentiate the wave function with respect to tt, then substitute x=2x = 2 and t=1t = 1.
  7. Percepatan Transversal di x=2x = 2 meter dan t=1t = 1 detik:

    • Differentiate the wave function twice with respect to tt, then substitute x=2x = 2 and t=1t = 1.
  8. Gambarkan Gelombang dengan Persamaan diatas (Graph the Wave Function):

    • Plot the function y(x,t)=32sin(4πx3πt+π)y(x, t) = \frac{3}{2} \sin(4\pi x - 3\pi t + \pi) for various tt values.

Let me know if you want me to go into detailed calculations for any of these parts!


Additional Questions

  1. How do we calculate wave speed when given wave number kk and angular frequency ω\omega?
  2. What physical significance does the initial phase π\pi have on the wave?
  3. How does changing PP affect the wave’s frequency and amplitude?
  4. What methods are there to find the transverse velocity at a given point and time?
  5. Why is the maximum transverse velocity dependent on ω\omega and amplitude?

Tip

When analyzing wave equations, remember that terms involving xx relate to spatial properties (like wavelength), while terms involving tt relate to temporal properties (like frequency and period).

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Math Problem Analysis

Mathematical Concepts

Wave equations
Trigonometric functions
Wave speed
Transverse velocity and acceleration

Formulas

Wave function: y(x, t) = (1/2)P sin(4πx − Pπt + π)
Phase of the wave: Φ = 4πx - Pπt + π
Wave speed: v = ω / k
Maximum transverse velocity: v_max = Aω
Maximum transverse acceleration: a_max = Aω^2

Theorems

Wave propagation theory

Suitable Grade Level

Undergraduate level