PHSX 220 Physics I (w/ Calculus) · Unit 4: Oscillations, waves, sound · on Final (Exam 4)
Ch 16 Mechanical waves
Study notes for Mechanical waves in PHSX 220 Physics I (w/ Calculus) at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in Mathub.
Big ideas
- A wave carries energy, not matter. Transverse (string) vs longitudinal (sound). Wavelength $\lambda$, period $T$, frequency $f = 1/T$, speed $v = \lambda f = \lambda/T$. The speed is set by the medium; the frequency by the source.
- Wave function $y(x,t) = A\sin(kx - \omega t + \phi)$ with wave number $k = 2\pi/\lambda$ and $\omega = 2\pi f$; speed $v = \omega/k$. The minus sign means motion in $+x$.
- String: $v = \sqrt{F_T/\mu}$ ($\mu$ = mass per length). Tighter or lighter strings carry faster waves. Wave power $\propto A^2 f^2$.
- Superposition: waves add. In phase → constructive; half a wavelength out of phase → destructive. Reflection from a fixed end inverts the pulse.
- Standing waves on a string fixed at both ends: $\lambda_n = 2L/n$, $f_n = n\dfrac{v}{2L}$ ($n = 1,2,3,\dots$). Nodes are $\lambda/2$ apart; the fundamental has one antinode.
Key formulas
Wave relations
$$v = \lambda f = \frac{\omega}{k},\quad k = \frac{2\pi}{\lambda},\quad \omega = 2\pi f,\quad v_{\text{string}} = \sqrt{\frac{F_T}{\mu}}$$
Standing waves (fixed–fixed)
$$\lambda_n = \frac{2L}{n},\qquad f_n = n\frac{v}{2L} = nf_1$$
Worked example
A 0.60 m guitar string (mass 2.4 g) is under 80 N of tension. Find the wave speed and the fundamental frequency.
Show the solution
$\mu = 0.0024/0.6 = 0.004$ kg/m; $v = \sqrt{80/0.004} = 141$ m/s; $f_1 = v/(2L) = 141/1.2 = 118$ Hz.
Common mistakes
- Thinking a louder source makes a faster wave. Speed is a property of the medium.
- Using the string length as the wavelength of the fundamental (it is half a wavelength).
Exam tipDraw the standing wave pattern: count the loops. n loops means λ = 2L/n and f = n f₁.
Keep going in Mathub
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