Exercise 10.11
Two friends are standing along a steel fence at a distance of m from each other (Fig. ). Gunjan places her ear over the fence and her friend knocks the fence with a metal object. Using the values of the speed of sound in steel and air given in Table , calculate the time difference between the sound that reached Gunjan through the air and the steel. Would it have been possible for her to distinguish between the two sounds? (The time interval between two sounds must be at least s to be heard separately.) Human perception of sound The physical properties of sound, such as time period, wavelength, frequency, amplitude and speed are well-defined and can be measured. However, how we experience sound is subjective. Human perception of sound is described by terms, such as loudness and pitch which do not have simple relations with the physical properties. Pitch How frequency is perceived by humans is called pitch. Sounds perceived to be shrill, such as a whistle or a siren are said to have high pitch, while deep sounds like thunder or an aircraft rumble have low pitch. In general, high pitch sounds have higher frequency and low pitch sounds have lower frequency, although the exact mathematical relation is complicated. Threads of Curiosity Everyone’s voice sounds unique and you can often instantly recognise your teacher or friends calling out your name. Why do our voices sound different? Male, female, and children’s voices differ not just in their frequency but also on how the sound is shaped by the throat, mouth, and nasal cavities. During adolescence, boys’ vocal cords of boys lengthen and thicken, vibrating less frequently thus, ‘deepening’ their voice. Let us carry out an activity to listen and appreciate sounds at different frequencies. Activity : Let us experiment (demonstration activity) This activity is recommended to be performed as a classroom group activity facilitated by teacher. 1. Open a mobile app that can generate sounds. 2. Set the frequency to Hz, tap ‘play’, and listen carefully. 3. Increase the frequency in steps of Hz up to Hz and describe how the sound changes. 4. Next, set the frequency to Hz. Reduce the frequency till about Hz or the point where you cannot hear the sound anymore.
NCERT’s answer
0.$\displaystyle 932$ s; Yes
Time difference \(\displaystyle = 0.932\ \text{s} \), and yes — Gunjan could hear the two as separate sounds.
Through air: \(\displaystyle t_{\text{air}} = \dfrac{d}{v_{\text{air}}} = \dfrac{340\ \text{m}}{340\ \text{m s}^{-1}} = 1\ \text{s} \).
Through steel: \(\displaystyle t_{\text{steel}} = \dfrac{340\ \text{m}}{5000\ \text{m s}^{-1}} = 0.068\ \text{s} \).
\(\displaystyle \Delta t = 1\ \text{s} - 0.068\ \text{s} = 0.932\ \text{s} \).
\(\displaystyle 0.932\ \text{s} \) is far more than the \(\displaystyle 0.1\ \text{s} \) the ear needs, so the two arrivals are heard separately.
The knock reaches her through the steel first (sound is fastest in solids), and the same knock arrives through the air nearly a second later.