This checkpoint covers Lessons 9 to 13: sound as a mechanical wave, waveforms, intensity, decibels, beats, standing waves in pipes, the Doppler effect, and evidence that sound is a wave.
Checkpoint Assessment
1. Sound in air is best described as a:
2. If the distance from a point sound source doubles, the intensity becomes:
3. A 10 dB increase means the sound is:
4. Two notes of 512 Hz and 516 Hz played together produce a beat frequency of:
5. A closed pipe differs from an open pipe because it:
6. A source moving towards an observer causes the observed frequency to:
7. Which provides evidence for sound interference?
8. Which statement correctly distinguishes sound from light?
9. Explain why the bell-jar experiment supports the claim that sound is a mechanical wave. 3 MARKS
10. An open pipe has length 0.60 m. Find the wavelength of its fundamental. 3 MARKS
11. A 700 Hz siren moves towards a stationary observer at 20 m/s. Take the speed of sound as 340 m/s. Find the observed frequency and explain why it is higher than the source frequency. 4 MARKS
1. Bsound in air is longitudinal and mechanical.
2. Ddoubling distance reduces intensity to one quarter.
3. Aa 10 dB increase means 10 times greater intensity.
4. C$|512 - 516| = 4\ \text{Hz}$.
5. Ba closed pipe has a node at the closed end and supports odd harmonics only.
6. Aapproaching motion raises observed frequency.
7. Dcoherent speakers can create interference maxima and minima.
8. Bsound needs a medium while light does not.
Q9 (3 marks): In the bell-jar experiment, the source continues vibrating while the sound fades as air is removed. This shows that sound needs a medium to be transmitted. Because it requires particles in a medium, sound is a mechanical wave.
Q10 (3 marks): For the fundamental of an open pipe, $L = \lambda/2$. So $\lambda = 2L = 1.2\ \text{m}$.
Q11 (4 marks): Use $f' = f \times \dfrac{v}{v - v_s}$. So $f' = 700 \times \dfrac{340}{340 - 20} = 700 \times \dfrac{340}{320} \approx 744\ \text{Hz}$. It is higher than the source frequency because the approaching source compresses the wavefronts, reducing the observed wavelength. With the same wave speed in air, this gives a higher observed frequency.
Tick when you have finished the checkpoint and checked the answers.