NCERT Solutions: Sound Waves

1. Ways of Producing Sound: Sound is produced by vibrating objects. Common methods include:

• Plucking stretched strings (e.g., guitar, violin).
• Blowing air columns into hollow pipes (e.g., flute, trumpet).
• Striking stretched membranes (e.g., tabla, drum).
• Striking metal or wooden plates/bars (e.g., tuning fork, xylophone, cymbals/taal).
• Vibrating vocal cords by forcing air from lungs through larynx.

2. Table of Musical Instruments & Vibrating Parts:

Sound is a mechanical wave that propagates by creating density/pressure disturbances in medium particles. In the bell jar experiment, as air is evacuated, there are fewer medium particles to transmit vibrations. In a near vacuum, sound cannot propagate at all. Hence, Reason (R) correctly explains Assertion (A).

Explanation: Assertion (A) is TRUE because compressions (high density) and rarefactions (low density) propagate forward as a disturbance. However, Reason (R) is FALSE because individual medium particles do NOT move continuously forward; they only oscillate back and forth about their fixed mean positions.

In wave propagation, it is the energy of the disturbance that travels through particle-to-particle collisions. The air particles themselves near the tuning fork stay in their local vicinity and oscillate back and forth.

Solution to Q8:

Frequency $\nu = 20\text{ Hz} = 20\text{ oscillations/second}$.

Number of oscillations in 1 minute ($60\text{ seconds}$) $= 20 \times 60 = \mathbf{1200\text{ oscillations/min}}$.

Solution to Q9:

From graph Fig. 10.19, the full wavelength $\lambda$ (distance from 0 to peak to next peak) $= 3.0\text{ cm}$.

Half wavelength $\frac{\lambda}{2} = \frac{3.0\text{ cm}}{2} = \mathbf{1.5\text{ cm}}$.

A mechanical wave is defined as a wave that propagates through the vibration of material particles and strictly requires a physical medium (solid, liquid, or gas) to travel. Electromagnetic waves (like light) also show reflection, carry energy, and have frequency without needing a medium. Therefore, the requirement of a material medium is the definitive proof that sound is a mechanical wave.

Frequency ($\nu$) is defined as the number of complete oscillations (compressions and rarefactions passing a fixed point) per unit time (second). Increasing frequency means more compressions pass per second.

Note: Speed $v$ depends only on the medium. Since $v = \lambda \nu$, increasing frequency decreases wavelength $\lambda$ and decreases time period $T = 1/\nu$.

Given: Number of compressions $N = 20$, Time $t = 4\text{ seconds}$.

Answer: It will produce reverberation.

Justification:

• The human brain retains the sensation of any sound for approximately $0.1\text{ seconds}$ (known as persistence of hearing).
• To hear a distinct, separate echo, the reflected sound must arrive at the ear at least $0.1\text{ s}$ after the original sound.
• Here, the reflected sound arrives in $0.05\text{ s}$, which is less than $0.1\text{ s}$ ($0.05\text{ s} < 0.1\text{ s}$).
• Therefore, the original sound and reflected sound overlap in the ear, resulting in a prolonged persistence of sound known as reverberation rather than a distinct echo.

Comparing graph (a) and graph (b) given in Fig. 10.30 on identical X and Y axes scales:

(i) Greater Wavelength ($\lambda$):

• Wavelength is the horizontal distance between two consecutive crests or troughs along the X-axis.
• In Graph (a), the crests are spread farther apart (fewer wave cycles per unit distance), meaning it has a larger crest-to-crest distance.
• Hence, Graph (a) has the greater wavelength.

(ii) Smaller Amplitude ($A$):

• Amplitude is the vertical height of a crest above the average baseline along the Y-axis.
• In Graph (a), the height of the wave peaks is visually smaller than the tall peaks in Graph (b).
• Hence, Graph (a) has the smaller amplitude.

Frequency corresponds to the number of wave cycles per unit distance on the graph:

• Green Curve: Shows the highest number of oscillations/cycles within the given distance $\implies$ Maximum Frequency $\implies$ Source A.
• Red Curve: Shows intermediate number of oscillations $\implies$ Medium Frequency $\implies$ Source B.
• Blue Curve: Shows the least number of oscillations (widest wavelength) within the given distance $\implies$ Minimum Frequency $\implies$ Source C.

Graph Specifications:

• Y-axis (Density Variation): Amplitude $A = 3\text{ units}$. Crest peak reaches $+3$, trough reaches $-3$ relative to baseline.
• X-axis (Distance in cm): Wavelength $\lambda = 4\text{ cm}$. One full cycle starts at $0\text{ cm}$, crest at $1\text{ cm}$, baseline crossing at $2\text{ cm}$, trough at $3\text{ cm}$, and completes at $4\text{ cm}$. The second cycle completes at $8\text{ cm}$.

There are two major scientific errors in this movie depiction:

1. Sound cannot travel in outer space (Vacuum): Outer space is a near vacuum with no material medium (air). Sound is a mechanical wave that requires a material medium to propagate. Therefore, an explosion in space produces no sound that can be heard by an external observer.
2. Speed difference between Light and Sound: Light travels at an extremely high speed ($c = 3 \times 10^8\text{ m/s}$), whereas sound in air travels much slower ($\approx 340\text{ m/s}$). Even if air were present, the flash of light would be seen almost instantly, while the sound would arrive much later—never at the exact same time.

Given:

• Wavelength ($\lambda$) $= 3.44\text{ m}$
• Speed of wave ($v$) $= 344\text{ m s}^{-1}$

Step 1: Calculate Frequency ($\nu$)

Step 2: Calculate Time Period ($T$)

Given:

• Total round-trip echo time ($t$) $= 5\text{ s}$
• Speed of ultrasound in seawater ($v$) $= 1525\text{ m s}^{-1}$

Formula:

In echo-ranging, the ultrasound signal travels down to the wreckage and reflects back to the ship, covering a total distance of $2d$.

Calculation:

Given:

• Distance to obstacle ($d$) $= 1.2\text{ m}$
• Speed of ultrasonic wave in air ($v$) $= 345\text{ m s}^{-1}$

Calculation:

Total round-trip distance traveled by ultrasonic wave to obstacle and back $= 2d = 2 \times 1.2\text{ m} = 2.4\text{ m}$.

Converting to milliseconds: $t \approx 0.006956 \times 1000 = \mathbf{6.96\text{ ms}}$.

Given:

• Distance ($d$) $= 1720\text{ m}$
• Speed at $22^{\circ}\text{C}$ ($v_1$) $= 344\text{ m s}^{-1}$
• Speed at $0^{\circ}\text{C}$ ($v_2$) $= 331\text{ m s}^{-1}$

Step 1: Time taken at $22^{\circ}\text{C}$ ($t_1$)

Step 2: Time taken at $0^{\circ}\text{C}$ ($t_2$)

Step 3: Calculate Extra Time ($\Delta t$)

Given:

• Speed of wave ($v$) $= 340\text{ m s}^{-1}$
• From Fig. 10.32: The bracket length of $8\text{ cm}$ spans 2 complete compressions/wavelengths ($2\lambda$).

Step 1: Calculate Wavelength ($\lambda$)

Step 2: Calculate Frequency ($\nu$)

Given: Speed of both waves $v = 345\text{ m s}^{-1}$.

Analysis for Wave A (Pink Curve):

• From graph Fig. 10.33, one complete wave cycle of A spans from $0$ to $2.5\text{ cm}$.
• Wavelength $\lambda_A = 2.5\text{ cm} = \mathbf{0.025\text{ metres}}$.
• Frequency $\nu_A = \frac{v}{\lambda_A} = \frac{345\text{ m s}^{-1}}{0.025\text{ m}} = \mathbf{13,800\text{ Hz}} = \mathbf{13.8\text{ kHz}}$.

Analysis for Wave B (Blue Curve):

• From graph Fig. 10.33, one complete wave cycle of B spans from $0$ to $5.0\text{ cm}$.
• Wavelength $\lambda_B = 5.0\text{ cm} = \mathbf{0.05\text{ metres}}$.
• Frequency $\nu_B = \frac{v}{\lambda_B} = \frac{345\text{ m s}^{-1}}{0.05\text{ m}} = \mathbf{6900\text{ Hz}} = \mathbf{6.9\text{ kHz}}$.

Given:

• Distance to cliff $d$ is identical for both sources A and B.
• Round-trip distance $= 2d$.
• Time taken in air ($t_A$) is $4.5$ times time taken in water ($t_B$): $t_A = 4.5 \times t_B$.

Mathematical Derivation:

Distance formula: $2d = v_{\text{air}} \times t_A$ and $2d = v_{\text{water}} \times t_B$.

Equating the two distances:

Rearranging for the speed ratio $\frac{v_{\text{air}}}{v_{\text{water}}}$: