MP Board · Class 9 · Science · Sound(a) Explain the fundamental characteristics of a sound wave: Amplitude, Wavelength, Frequency, Time Period, and Wave Speed. State clearly how the pitch and loudness of a sound depend on these characteristics. (b) Numerical Problem: A bat emits an ultrasonic sound of frequency $100\text{ kHz}$ in air. If this sound meets a water surface, calculate the wavelength of: (i) the reflected sound wave in air (speed of sound in air = $340\text{ m/s}$) (ii) the transmitted sound wave in water (speed of sound in water = $1486\text{ m/s}$)

(a) Fundamental Characteristics of a Sound Wave

\nA sound wave is a longitudinal mechanical wave propagating through medium oscillations. Its key characteristics are:

1. Wavelength ($\lambda$):

   • The distance between two consecutive compressions or two consecutive rarefactions is called wavelength.
   • Its SI unit is meter (m).

2. Amplitude ($A$):

   • The magnitude of the maximum disturbance or displacement of medium particles from their mean position is called amplitude.
   • It determines the energy carried by the wave.

3. Frequency ($f$ or $\nu$):

   • The number of complete wave oscillations per unit time is called frequency.
   • Its SI unit is Hertz (Hz).

4. Time Period ($T$):

   • The time required for one complete oscillation or cycle of the density variation in the medium.
   • $T = \frac{1}{f}$. Its SI unit is second (s).

5. Wave Speed ($v$):

   • The distance traveled by a sound wave per unit time.
   • Relationship formula: $\text{Speed } (v) = \text{Frequency } (f) \times \text{Wavelength } (\lambda)$.

Dependence of Pitch and Loudness:

• Pitch: Pitch depends directly on the frequency of the sound wave. A higher frequency produces a sound with a higher pitch (shrill sound), whereas a lower frequency results in a lower pitch (grave/deep sound).
• Loudness: Loudness depends on the amplitude of the vibrating body. Loudness is proportional to the square of the amplitude ($\text{Loudness} \propto A^2$). Larger amplitude produces a louder sound, while smaller amplitude produces a soft sound.

(b) Solution to Numerical Problem

Given:

• Frequency of ultrasound ($f$) = $100\text{ kHz} = 100,000\text{ Hz} = 10^5\text{ Hz}$
• Speed of sound in air ($v_{\text{air}}$) = $340\text{ m/s}$
• Speed of sound in water ($v_{\text{water}}$) = $1486\text{ m/s}$

(Note: Frequency of a wave remains constant regardless of reflection or change in medium.)

(i) Wavelength of reflected sound wave in air ($\lambda_{\text{air}}$): $$\lambda_{\text{air}} = \frac{v_{\text{air}}}{f}$$ $$\lambda_{\text{air}} = \frac{340\text{ m/s}}{10^5\text{ Hz}} = 3.4 \times 10^{-3}\text{ m} = 3.4\text{ mm}$$

(ii) Wavelength of transmitted sound wave in water ($\lambda_{\text{water}}$): $$\lambda_{\text{water}} = \frac{v_{\text{water}}}{f}$$ $$\lambda_{\text{water}} = \frac{1486\text{ m/s}}{10^5\text{ Hz}} = 1.486 \times 10^{-2}\text{ m} = 14.86\text{ mm}$$

Answer:

• Wavelength of reflected sound in air = $3.4 \times 10^{-3}\text{ m}$ ($3.4\text{ mm}$).
• Wavelength of transmitted sound in water = $1.486 \times 10^{-2}\text{ m}$ ($14.86\text{ mm}$).