Interference of Waves

What this figure shows. Two wave pulses of the same shape are shown approaching each other and overlapping, with their combined displacement at the crossing point drawn as visibly TALLER than either individual pulse -- the algebraic sum of the two, since both pulses are displaced in the same direction at that point. The figure gives the first concrete visual instance of interference: when two waves of matching shape and phase overlap, their displacements simply add, producing a combined disturbance who …

What this figure shows. Two sinusoidal waves, labelled $y_1$ and $y_2$, are drawn on the same axes with a visible horizontal offset between them corresponding to a phase difference marked as $\phi=60^{\circ}$, and a third curve labelled y shows their sum at every point along the x-axis. The resulting summed curve y is seen to still be a smooth sinusoid of the same frequency as the two originals, but with its own amplitude and its own phase, lying somewhere between the two input curves. The figure is the graphical companion to the algebraic derivation in the text: it shows visually that adding two same-frequency sinusoids with a phase offset between them always produces a third sinusoid of that same frequency, whose amp …

What this figure shows. Panel (a) shows two identical wave curves, labelled Wave 1 and Wave 2, drawn one above the other with their crests and troughs perfectly aligned, and a plus sign and equals sign lead to a third, resultant curve drawn noticeably taller than either input wave -- their crests have added together. Panel (b) shows the same two waves instead drawn with crests of one aligned against troughs of the other, and the resultant curve after the plus and equals signs is drawn as a flat, nearly zero-amplitude line, since the two displacements cancel at every point. The figure gives the clearest possible visual summary of the section's two central outcomes: interference is constructive (louder, amplified) when the two waves are in phase (aligned crest-to-crest), and destructive (q …

What this figure shows. A T-shaped tube apparatus is drawn with a loudspeaker S feeding sound into a junction point P, from which the tube splits into two separate paths -- one of fixed length labelled $r_1$, and a second, U-shaped sliding tube of adjustable length labelled $r_2$ -- that recombine at a receiver R. The figure sets up the experimental method used in the worked examples that follow: because sound entering at P splits in two, travels the two different path lengths $r_1$ and $r_2$, and recombines at R, sliding the adjustable tube to change $r_2$ directly changes the path difference $\Delta r=|r_2-r_1|$ between the two arriving waves, letting a listener at R be swept smoothly through alternating loud (constructive) and quiet (destructive) points as the path differen …

What this figure shows. Two identical sinusoidal curves, labelled $y_1$ and $y_2$, are drawn perfectly overlapping each other along the same x-axis, since their phase difference is exactly $\phi=0^{\circ}$, illustrating two waves arriving at a receiver completely in step with one another. Because the two curves coincide exactly crest-for-crest and trough-for-trough, their sum doubles the displacement at every point, giving the maximum possible resultant intensity -- this is the visual case of the receiver R, in the apparatus of Figure 11.30, positioned where the path difference between t …

What this figure shows. Two sinusoidal curves, labelled $y_1$ and $y_2$, are drawn shifted exactly half a cycle apart along the x-axis, so that wherever $y_1$ has a crest, $y_2$ has a trough at the very same point, corresponding to a phase difference of $\phi=180^{\circ}$. Because the two curves are everywhere equal and opposite in sign, their sum cancels to (near) zero at every point, giving the minimum possible resultant intensity -- this is the visual case of the receiver R, in the apparatus of Figure 11.30, positioned where the path difference between the two arms is a half-odd-integer number of wavelengths …

Worked out. Two sources A and B emit identical-frequency waves in the same phase, and the task is to find the resultant intensity at three points: O (equidistant from A and B), Y (where the path difference is exactly one wavelength lambda), and X (where the path difference is exactly half a wavelength). At point O, since OA = OB the path difference is zero, so the two waves arrive in phase and interfere constructively, giving maximum intensity. At point Y, a path difference of $\Delta r=\lambda$ corresponds to a phase difference $\Delta\phi=(2\pi/\lambda)\times\lambda=2\pi$, which is a full cycle and therefore also equivalent to being in phase, so Y also shows maximum intensity. At point X, a path difference of $\Delta r=\lambda/2$ corresponds to $\Delta\phi=(2\pi/\lambda)\times(\lambda/2)=\pi$, exactly out of phase, so the waves interfere destructively at X and t …

Worked out. Two speakers C and E, 5 m apart and driven by the same source, sit with a man initially standing at point A, 10 m from the midpoint O of C and E; walking toward O along a line 1 m away from and parallel to OC, he detects the first minimum in sound intensity at a point B, and the task is to find the driving frequency, given the speed of sound as 343 m/s. First-minimum means the two path lengths from C and E to B differ by exactly half a wavelength, $\Delta x=\lambda/2$. Using right-triangle geometry (drawing perpendiculars from C and E onto the line through A and B) to work out the two path lengths $x_1$ (from the nearer speaker) and $x_2$ (from the farther speaker) gives, after computing each hypotenuse from the known 10 m and the relevant 1.5 m or 3.5 m offsets, $x_1\approx10.1\ \text{m}$ and $x_2\approx10.6\ \text{m}$, so the path difference is $\Delta x = x_2-x_1\approx0.5\ \text{m}$. Setting this equal to $\lambda/2$ gives $\lambda=1.0\ \text{m}$, and the fr …