We are given the wave disturbance:
y(x,t)=0.02cos(50πt+π2)cos(10πx)y(x,t) = 0.02 \cos \left( 50\pi t + \frac{\pi}{2} \right) \cos(10\pi x)
Step 1: Understanding the wave equation
The general form of a wave equation can be written as:
y(x,t)=Acos(ωt+ϕ)cos(kx)y(x,t) = A \cos(\omega t + \phi) \cos(kx)
Where:
• AA is the amplitude of the wave,
• ω\omega is the angular frequency of the wave,
• ϕ\phi is the phase constant,
• kk is the wave number,
• xx is the position, and
• tt is the time.
Step 2: Comparing the given wave equation with the standard form
In our equation:
y(x,t)=0.02cos(50πt+π2)cos(10πx)y(x,t) = 0.02 \cos \left( 50\pi t + \frac{\pi}{2} \right) \cos(10\pi x)
• The angular frequency ω=50π\omega = 50\pi rad/s.
• The wave number k=10πk = 10\pi rad/m.
Step 3: Calculating the wavelength and wave speed
• The angular frequency ω\omega is related to the frequency ff by the equation ω=2πf\omega = 2\pi f. Thus, the frequency ff is:
f=ω2π=50π2π=25 Hzf = \frac{\omega}{2\pi} = \frac{50\pi}{2\pi} = 25 \, \text{Hz}
• The wave number kk is related to the wavelength λ\lambda by the equation k=2πλk = \frac{2\pi}{\lambda}. Thus, the wavelength λ\lambda is:
λ=2πk=2π10π=0.2 m\lambda = \frac{2\pi}{k} = \frac{2\pi}{10\pi} = 0.2 \, \text{m}
• The wave speed vv is related to the frequency and wavelength by the equation v=fλv = f\lambda. Therefore:
v=25×0.2=5 m/sv = 25 \times 0.2 = 5 \, \text{m/s}
Step 4: Identifying nodes and antinodes
• A node occurs where cos(kx)=0\cos(kx) = 0, which means kx=π2,3π2,5π2,…kx = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots. Therefore, the nodes occur at positions where:
x=(2n+1)π2kforn=0,1,2,…x = \frac{(2n+1)\pi}{2k} \quad \text{for} \quad n = 0, 1, 2, \dots
• An antinode occurs where cos(kx)=±1\cos(kx) = \pm 1, which means kx=nπkx = n\pi. Therefore, the antinodes occur at positions where:
x=nπkforn=0,1,2,…x = \frac{n\pi}{k} \quad \text{for} \quad n = 0, 1, 2, \dots
Step 5: Answering the options
Option A: A node occurs at x=0.15 mx = 0.15 \, \text{m}
• To check if there is a node at x=0.15 mx = 0.15 \, \text{m}, we substitute x=0.15x = 0.15 into the equation for nodes:
kx=10π×0.15=1.5πkx = 10\pi \times 0.15 = 1.5\pi
Since 1.5π1.5\pi is an odd multiple of π2\frac{\pi}{2}, there will be a node at x=0.15 mx = 0.15 \, \text{m}. This is correct.
Option B: An antinode occurs at x=0.3 mx = 0.3 \, \text{m}
• To check if there is an antinode at x=0.3 mx = 0.3 \, \text{m}, we substitute x=0.3x = 0.3 into the equation for antinodes:
kx=10π×0.3=3πkx = 10\pi \times 0.3 = 3\pi
Since 3π3\pi is an integer multiple of π\pi, there will be an antinode at x=0.3 mx = 0.3 \, \text{m}. This is correct.
Option C: The speed of the wave is 5 m/s5 \, \text{m/s}
• We already calculated the speed of the wave as 5 m/s5 \, \text{m/s}. This is correct.
Option D: The wavelength is 0.3 m0.3 \, \text{m}
• We already calculated the wavelength as 0.2 m0.2 \, \text{m}, not 0.3 m0.3 \, \text{m}. This is incorrect.
Final Answer:
The correct options are:
• A. A node occurs at x=0.15 mx = 0.15 \, \text{m}
• B. An antinode occurs at x=0.3 mx = 0.3 \, \text{m}
• C. The speed of the wave is 5 m/s5 \, \text{m/s}
