Waves: Numerical Problems Set (1)

easy • 3 min read
Tags Waves

Question

A transverse wave on a stretched string has the equation y(x,t)=0.05sin⁡(20πt−4πx)y(x,t) = 0.05 \sin(20\pi t - 4\pi x) where xx and yy are in metres and tt in seconds. Find: (a) amplitude, (b) wavelength, (c) frequency, (d) wave speed, (e) maximum particle speed.

Solution — Step by Step

Standard form: y=Asin⁡(ωt−kx)y = A\sin(\omega t - kx). By inspection:

A=0.05 mA = 0.05 \text{ m}, ω=20π rad/s\omega = 20\pi \text{ rad/s}, k=4π rad/mk = 4\pi \text{ rad/m}.

 λ=2πk=2π4π=0.5 m\lambda = \tfrac{2\pi}{k} = \tfrac{2\pi}{4\pi} = 0.5 \text{ m} f=ω2π=20π2π=10 Hzf = \tfrac{\omega}{2\pi} = \tfrac{20\pi}{2\pi} = 10 \text{ Hz}

Two ways to compute it:

v=fλ=10×0.5=5 m/sorv=ωk=20π4π=5 m/sv = f\lambda = 10 \times 0.5 = 5 \text{ m/s} \quad \text{or} \quad v = \tfrac{\omega}{k} = \tfrac{20\pi}{4\pi} = 5 \text{ m/s}

Particle velocity is ∂y∂t=Aωcos⁡(ωt−kx)\tfrac{\partial y}{\partial t} = A\omega \cos(\omega t - kx). Maximum value:

vp,max⁡=Aω=0.05×20π=π≈3.14 m/sv_{p,\max} = A\omega = 0.05 \times 20\pi = \pi \approx 3.14 \text{ m/s}

Final answers: A=0.05 mA = \mathbf{0.05 \text{ m}}, λ=0.5 m\lambda = \mathbf{0.5 \text{ m}}, f=10 Hzf = \mathbf{10 \text{ Hz}}, v=5 m/sv = \mathbf{5 \text{ m/s}}, vp,max⁡=π m/sv_{p,\max} = \mathbf{\pi \text{ m/s}}.

Why This Works

The wave equation y=Asin⁡(ωt−kx)y = A\sin(\omega t - kx) packs four pieces of information: amplitude (the coefficient), angular frequency (coefficient of tt), wave number (coefficient of xx), and direction (sign). Once you read these off, every other quantity is one short formula away.

The distinction between wave speed (v=ω/kv = \omega/k, the speed of the disturbance) and particle speed (AωA\omega, how fast individual points oscillate) is a classic trap. They’re different physical things.

Alternative Method

If the equation were given as y=Asin⁡ 2π(ft−x/λ)y = A\sin\,2\pi(ft - x/\lambda), you’d read off ff and λ\lambda directly. Both forms are equivalent — practise converting between them so neither version surprises you in the exam.

Common Mistake

Confusing wave speed and particle speed. The wave moves at 5 m/s5 \text{ m/s} but a single point on the string oscillates with maximum speed π≈3.14 m/s\pi \approx 3.14 \text{ m/s}. They are not the same and can have very different magnitudes.

To find the direction of propagation: if the equation is sin⁡(ωt−kx)\sin(\omega t - kx) the wave moves in +x+x; if it’s sin⁡(ωt+kx)\sin(\omega t + kx) the wave moves in −x-x. The relative sign of ωt\omega t and kxkx tells the story.

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