MP Board · Class 9 · Science · Sound(a) Explain how a sound wave propagates through a medium as a series of compressions and rarefactions. (b) A SONAR system installed on an ocean research vessel emits an ultrasonic signal towards the seabed. The echo is detected back after a time interval of $3.6\text{ seconds}$. If the speed of sound in seawater is $1531\text{ m/s}$, calculate: The depth of the seabed below the vessel. The wavelength of the ultrasonic signal in seawater, given that its frequency is $50\text{ kHz}$.

(a) Propagation of Sound as Compressions and Rarefactions

• Sound requires a material medium (solid, liquid, or gas) for its propagation.
• When a vibrating object moves forward, it pushes and compresses the air in front of it, creating a region of high pressure and high density called a Compression (C).
• When the vibrating body moves backward, it creates a region of low pressure and low density called a Rarefaction (R).
• As the object vibrates rapidly to and fro, a continuous series of compressions and rarefactions is produced in the air.
• These pressure variations propagate through the medium. It is important to note that individual particles of the medium do not travel all the way from the source to the listener; they merely oscillate back and forth about their equilibrium positions, transferring energy to adjacent particles.

(b) Numerical Problem Working

Part 1: Calculation of Depth of the Seabed

Given:

• Total time elapsed before echo is received ($t$) = $3.6\text{ s}$
• Speed of sound in seawater ($v$) = $1531\text{ m/s}$

Formula:\nLet the depth of the seabed be $d$.\nThe total distance traveled by the sound wave going to the seabed and returning back is $2d$. $$\text{Total Distance} = \text{Speed} \times \text{Time}$$ $$2d = v \times t$$

Calculation: $$2d = 1531\text{ m/s} \times 3.6\text{ s}$$ $$2d = 5511.6\text{ m}$$ $$d = \frac{5511.6}{2} = 2755.8\text{ m}$$

Hence, the depth of the seabed is $2755.8\text{ meters}$ (or $2.756\text{ km}$).

Part 2: Calculation of Wavelength of the Sound Wave

Given:

• Frequency of sound wave ($\nu$) = $50\text{ kHz} = 50 \times 10^3\text{ Hz} = 50,000\text{ Hz}$
• Speed of sound wave ($v$) = $1531\text{ m/s}$

Formula: $$\text{Speed of wave } (v) = \text{Frequency } (\nu) \times \text{Wavelength } (\lambda)$$ $$\lambda = \frac{v}{\nu}$$

Calculation: $$\lambda = \frac{1531}{50000}$$ $$\lambda = 0.03062\text{ m} = 3.062\text{ cm}$$

Hence, the wavelength of the ultrasonic signal in seawater is $0.03062\text{ m}$ (or $3.062\text{ cm}$).