EduNinja

(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 .

[ 3 ]

(i)

Outline how the standing wave is formed in the pipe.

[ 2 ]

(ii)

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 solid line represents the standing wave at time t and the dotted line represents the standing wave at an instant later. The dot is the equilibrium position of a particle P in the pipe. The up arrow indicates displacements to the right and the down arrow displacements to the left.

On the diagram, draw

[ 2 ]

(i)

a dot to indicate the approximate position of P at time t,

[ 1 ]

(ii)

an arrow to indicate the velocity of P at time t.

[ 1 ]

(c)

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]

IB Physics HL C 4 Standing Waves and Resonance Topic Practice

Learn this topic

IB Physics HL C 4 Standing Waves and Resonance Topic Practice

Learn this topic

Question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)

Question (c)