IB Physics HL C.4.3 Strings and Pipes Question Bank

Practise IB Physics HL C.4.3 by solving string and pipe harmonic spectra, mixed-boundary conditions, wave-speed and resonance problems.

[Maximum number: 3]

(a)

A loudspeaker emits sound of frequency 210 Hz into a pipe with one open and one closed end. The diagram shows a representation of the standing wave established in the pipe.

The length of the pipe is 1.20 m .

[ 1 ]

(i)

Determine the wavelength of the wave.

[ 1 ]

$\lambda=\lessdot \frac{4 L}{3}=\frac{4 \times 1.20}{3}=>1.6<\mathrm{m} \gg$

[1]

(b)

The frequency of sound is reduced to 140 Hz . Explain why a standing wave will not be formed in the pipe.

[ 2 ]

ALTERNATE 1

the pipe can only support standing waves with frequencies that are odd multiples of the first harmonic frequency

first harmonic frequency is $70 \mathrm{~Hz}$

ALTERNATE 2

<<the new wavelength would be 2.4 m so>> a node would be formed at the open end

An antinode is required at the open end to form a standing wave

[2]

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C.4.3—Strings and pipes question 1

Question (a)

Question (i)

Question (b)