23P2Q04

4a)

The incident sound wave reflects off the metal plate.

The incident and reflected sound waves, being sound waves of the same frequency travelling in opposite directions, superpose to form a stationary wave.

4aii)

One complete cycle has a width of 5 cm on the oscilloscope.

Period \displaystyle T=5\times 0.50=2.5\text{ ms}

Frequency \displaystyle f=\frac{1}{T}=\frac{1}{2.5\times {{10}^{-3}}}=400\text{ Hz}

4aiii)

\displaystyle \begin{aligned} (v&=f\lambda ) \\340&=(400)\lambda  \\  \lambda &=0.85\text{ m}  \end{aligned}

Distance between adjacent nodes\displaystyle =\frac{\lambda }{2}=0.85\div 2=0.425\text{ m}

4bi)

Path difference\displaystyle =YP-XP=1.8-1.4=0.4\text{ m}, which corresponds to \displaystyle 0.4\div 0.20=2\lambda .

Since path difference is an integer number of wavelengths, the two waves will meet in phase, under constructive interference and forms a maximum at P.

4bii)

\displaystyle \begin{aligned}  (I&\propto \frac{1}{{{r}^{2}}}) \\  \frac{{{I}_{Y}}}{{{I}_{X}}}&=\frac{X{{P}^{2}}}{Y{{P}^{2}}} \\   \frac{{{I}_{Y}}}{4.5\times {{10}^{-6}}}&={{(\frac{1.4}{1.8})}^{2}} \\   {{I}_{Y}}&=2.7\times {{10}^{-6}}\text{ W }{{\text{m}}^{-2}}  \end{aligned}

4biii)

\displaystyle I\propto {{A}^{2}}

\displaystyle \frac{{{A}_{X}}}{{{A}_{Y}}}=\sqrt{\frac{{{I}_{X}}}{{{I}_{Y}}}}=\sqrt{{{(\frac{1.8}{1.4})}^{2}}}=\frac{1.8}{1.4}=1.29

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