# IBDP Physics 4.5 – Standing waves: IB Style Question Bank – HL Paper 1

### Question

A metal rod of length 45 cm is clamped at its mid point. The speed of sound in the metal rod is $$1500\; ms^{-1}$$ and the speed of sound in air is $$300\; ms^{-1}$$. The metal rod vibrates at its first harmonic. What is the wavelength in air of the sound wave produced by the metal rod?

A 4.5 cm

B 9.0 cm

C 18 cm

D 90 cm

Ans: C

As rod is clamped at middle, its a node and both ends are antinodes

$$\therefore \frac{\lambda}{4}+\frac{\lambda}{4}=\frac{\lambda}{2}=L=0.45 \;m$$
or
$$\lambda =0.45 \times 2 =0.90$$
$$v=f \lambda$$
$$f= \frac{v}{\lambda} =\frac{1500}{0.90}$$
With same frequency air will also vibrate
$$\lambda_{air} =\frac{v_{air}}{f}=\frac{300}{\frac{1500}{0.90}}=\frac{0.96}{5}=0.18 \;m = 18\;cm$$

### Question

A string stretched between two fixed points sounds its second harmonic at frequency f.

Which expression, where n is an integer, gives the frequencies of harmonics that have a node at the centre of the string?

A. $$\frac{{n + 1}}{2}f$$

B. nf

C. 2nf

D. (2n + 1)f

### Markscheme

B

$$n\frac{\lambda}{2}=L$$
$$\therefore f_n =\frac{v}{\lambda} =\frac{nv}{2L}$$ —(1)
now for first Harmonic $$n=1$$
or
$$\lambda = 2L$$
$$\therefore f=\frac{v}{\lambda}=\frac{v}{2L}$$
putting this value in equation (1)
$$f_n =\frac{nv}{2L} =nf$$

### Question

The diagram shows a second harmonic standing wave on a string fixed at both ends.

What is the phase difference, in rad, between the particle at X and the particle at Y?

A. 0

B. $$\frac{\pi }{4}$$

C. $$\frac{\pi }{2}$$

D. $$\frac{{3\pi }}{4}$$

We have phase difference $$\phi$$ as
$$\phi = \frac{2\pi}{\lambda}\times \Delta x$$
$$\Delta x =x-y=0$$
$$\phi =0$$