A composite rod is formed by connecting two rods A and B as shown in the figure. Longitudinal wave velocities in rods A and B are 3210 m s⁻¹ and 6420 m s⁻¹, respectively. A longitudinal pulse applied at the free end of the rod A propagates with a wavelength of 2 m. What is the wavelength of this wave when it propagates through rod B?

2019 A/L Physics Past Paper · Question 16 (English Medium)

Wavelength of a longitudinal wave in a composite rod

A composite rod is formed by connecting two rods $A$ and $B$ as shown in the figure. Longitudinal wave velocities in rods $A$ and $B$ are $3210\ \text{m s}^{-1}$ and $6420\ \text{m s}^{-1}$ , respectively. A longitudinal pulse applied at the free end of the rod $A$ propagates with a wavelength of $2\ \text{m}$ . What is the wavelength of this wave when it propagates through rod $B$ ?

(1) 1 m
(2) 2 m
(3) 3 m
(4) 4 m
(5) 5 m

Answer and explanation

Answer: (4) 4 m

Given information

We know:

Rod $A$ : $v_A = 3210\ \text{m s}^{-1}$ .

Rod $B$ : $v_B = 6420\ \text{m s}^{-1}$ .

In rod $A$ the wavelength is $\lambda_A = 2$ m.

What the question asks

We want:

The wavelength $\lambda_B$ of the same wave in rod $B$ .

Relevant theory

The relationship among the wave velocity v, frequency f and the wavelength λ, can be written in the form of v = f λ.

This equation can be applied to all types of waves.

The equation v = fλ applies to the waves on each side of the boundary BD, where they have the same frequency, and from this it follows that their wavelengths are at the same ratio as their velocities.

In passing from one medium to another, waves undergo changes of speed and wavelength, but their frequency remains unchanged.

Required equations

$v = f\lambda$

$f = \frac{v_A}{\lambda_A}$

$\lambda_B = \frac{v_B}{f}$

Applying it to this question

The plan:

First find the frequency from rod $A$ .

The frequency does not change at the join.

Then use the speed in rod $B$ to find $\lambda_B$ .

Reasoning and calculation

$1605$

$f = \frac{v_A}{\lambda_A} = \frac{3210}{2}$

$\lambda_B = \frac{v_B}{f} = \frac{6420}{1605}$

$4$

$\lambda_B = 4$

$2$

$\frac{\lambda_B}{\lambda_A} = \frac{v_B}{v_A} = \frac{6420}{3210}$

$\lambda_B = 2 \times 2$

Checks

Check the unit.

$\dfrac{\text{m s}^{-1}}{\text{m}} = \text{s}^{-1}$ , that is Hz. This is a frequency.

$\dfrac{\text{m s}^{-1}}{\text{Hz}} = \text{m}$ . This is a wavelength.

Does it make sense?

In rod $B$ the wave is twice as fast.

The same number of bands arrive each second.

So they must be twice as far apart.

$\lambda_B = 2\lambda_A$ ✓.

$vB * lamA / vA$

Conclusion

$\lambda_B = 4\ \text{m}$

The frequency $1605$ Hz does not change:

$\lambda_B = \dfrac{6420}{1605} = 4\ \text{m}$

So the answer is option (4).

Taking the wavelength to stay the same and the frequency to change at the join.

The source decides the frequency. It does not change at the join. The speed and the wavelength change.

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