Class 10 Physics • Chapter 07 (Deep Detail)
Echo: The sound heard after reflection from a distant obstacle (like a cliff) after the original sound has ceased.
1. Minimum distance ($d$) between source and reflector should be 17 metres (in air).
2. Time interval between original and reflected sound must be at least 0.1 seconds.
Where $v$ = speed of sound, $d$ = distance, $t$ = time of echo.
NUMERICAL A ship sends a sonar signal and receives the echo after 4 seconds. If speed of sound in water is 1450 m/s, find depth of ocean.
Solution:
Total time $t = 4 \text{ s}$. Speed $v = 1450$.
$d = \frac{v \times t}{2} = \frac{1450 \times 4}{2}$
$= 1450 \times 2 = 2900 \text{ m}$.
NUMERICAL A person stands at distance $x$ from a cliff and claps. Echo is heard after 0.8 s. Find $x$. ($v_{air} = 340 \text{ m/s}$).
Solution:
$x = \frac{v \times t}{2} = \frac{340 \times 0.8}{2} = 170 \times 0.8 = 136 \text{ m}$.
Q: Can you hear an echo in a small room? Why?
Ans: No. Distance is less than 17 m, so reflected sound merges with original sound (Reverberation).
NUMERICAL From the displacement-distance graph shown above, find the Amplitude and Wavelength.
Solution:
1. Amplitude: Maximum displacement from mean position. Peak is at $y=5$. So, $A = 5 \text{ cm}$.
2. Wavelength ($\lambda$): Length of one complete wave cycles. Graph completes one cycle at $x=20$. So, $\lambda = 20 \text{ cm}$.
NUMERICAL A man stands between two cliffs and fires a gun. He hears the first echo after 1.5 s and the second echo after 2.5 s. If the speed of sound is 340 m/s, calculate the distance between the cliffs.
Solution:
Solution:
$d = d_1 + d_2 = \frac{v t_1}{2} + \frac{v t_2}{2}$
$= \frac{170 \times (1.5 + 2.5)}{1}$
$= 170 \times 4 = 680 \text{ m}$.
Vibrations of a body in the absence of any external resistive force. The body vibrates with its Natural Frequency and constant amplitude.
(Possible only in vacuum).
Vibrations in a medium (like air) where amplitude decreases with time due to friction/resistance. Finally, the body stops.
Vibrations of a body under the influence of an external periodic force. The body vibrates with the frequency of the applied force, not its natural frequency.
A special case of forced vibration when the Frequency of External Force = Natural Frequency of the body.
EXPERIMENT Four pendulums A, B, C, and D are suspended from the same elastic string. A and C are of the same length, while B is shorter and D is longer. If A is set into vibration, what do you observe?
Ans: Pendulum C will start vibrating with maximum amplitude because its length is equal to A, so its natural frequency matches A's frequency (Resonance). B and D will vibrate with small amplitudes (Forced Vibration).
CONCEPTUAL What is the essential condition for resonance?
Ans: The frequency of the driving force (external periodic force) must be exactly equal to the natural frequency of the vibrating body.
REASONING Why are stringed instruments provided with a hollow sound box?
Ans: The hollow box provides a large volume of air. The vibrating string causes the air to vibrate (Forced Vibration). The large surface area and resonance amplifies the sound.
CONCEPTUAL Two sounds have the same loudness and pitch but different quality. Draw their waveforms.
Ans: See diagram. One is a pure sine wave (monotone), the other is complex (rich in subsidiary notes). This difference in waveform gives them different quality.
Q: Who has a higher pitch voice: a mosquito or a lion?
Ans: Mosquito. (High frequency buzzing). Lion's roar is loud (High Amplitude) but low pitch.
NUMERICAL Two waves have amplitudes in the ratio 1:4. What is the ratio of their loudness (intensity)?
Solution: $I \propto A^2$. Ratio = $(1/4)^2 = 1:16$.
CONCEPTUAL How does the pitch of a stretched string depend on its length and tension?
Ans: $f \propto \frac{1}{L}$ and $f \propto \sqrt{T}$. Shorter string and Higher Tension $\implies$ Higher Pitch.
Loudness: Subjective. Depends on listener's ear sensitivity.
Intensity: Objective. Measurable quantity (Energy per unit area per second).