Class 10 - Unit # 11 : Sound - Solved Numericals

 

Problem 1

Calculate the speed of sound in air at 50°C. Given that the speed of sound at 0°C is 331 m/s. (Ans: 361 m/s or 361.5 m/s depending on formula used; 357.48 m/s using linear approximation)

Given Data:

• Speed of sound at 0°C (v_o) = 331 m/s
• Temperature (t) = 50°C
• Speed of sound at 50°C (v_t) = ?

Solution:

The speed of sound at any temperature t (in °C) can be calculated using the linear approximation formula:

v_t = v_o + 0.61t

Substituting the given parameters into the equation:

v_t = 331 + 0.61(50)

v_t = 331 + 30.5

v_t = 361.5 m/s

Note: If using the approximation constant 0.53:
v_t = 331 + 0.53(50) = 331 + 26.5 = 357.5 m/s.

Result:

The speed of sound in air at 50°C is approximately 361.5 m/s (or 357.5 m/s depending on the textbook's approximation constant).

Problem 2

A person has an audible range from 20 Hz to 20 kHz. What are the distinguishing wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 340 m/s. (Ans: Minimum λ = 17 mm, Maximum λ = 17 m)

Given Data:

• Minimum frequency (f_min) = 20 Hz
• Maximum frequency (f_max) = 20 kHz = 20,000 Hz
• Speed of sound (v) = 340 m/s
• Wavelength for minimum frequency (λ_max) = ?
• Wavelength for maximum frequency (λ_min) = ?

Solution:

Using the wave velocity relation v = f · λ, which rearranges to:

λ =

v

f

1. For Minimum Frequency (yielding Maximum Wavelength):

λ_max =

340

20

λ_max = 17 m

2. For Maximum Frequency (yielding Minimum Wavelength):

λ_min =

340

20000

λ_min = 0.017 m

λ_min = 17 mm

Result:

The wavelength corresponding to the lowest audible limit is 17 m, and the wavelength corresponding to the highest limit is 17 mm.

Problem 3

A ship uses ultrasonic pulses to measure the depth of a submarine beneath the ship. A sound pulse is transmitted into the sea, and the echo from the submarine is received after 40 ms. The speed of sound in seawater is 1480 m/s. Calculate the depth of the submarine. (Ans: 29.6 m)

Given Data:

• Round-trip echo time (t) = 40 ms = 40 × 10^-3 s = 0.04 s
• Speed of sound in seawater (v) = 1480 m/s
• Depth of the submarine (d) = ?

Solution:

First, calculate the total round-trip distance (S) traveled by the ultrasonic pulse:

S = v × t

S = 1480 × 0.04 = 59.2 m

Because the sound travels down to the submarine and reflects back, the actual depth (d) is exactly half of this total distance:

d =

S

2

=

59.2

2

d = 29.6 m

Result:

The depth of the submarine beneath the ship is 29.6 meters.

Problem 4

At night, bats emit pulses of sound to detect their prey. The speed of sound in air is 340 m/s.
(i) A bat emits a pulse of sound of wavelength 0.0080 m. Calculate the frequency of the sound.
(ii) The pulse of sound hits its prey and is reflected back to the bat. The bat receives the pulse 0.10 s after it is emitted. Calculate the distance traveled by the pulse of sound during this time.
(iii) Calculate the distance of the prey from the bat. (Ans: (i) 42500 Hz, (ii) 34 m, (iii) 17 m)

Given Data:

• Speed of sound in air (v) = 340 m/s
• Wavelength (λ) = 0.0080 m
• Time delay of echo (t) = 0.10 s

Solution:

(i) Calculate the Frequency of the Sound (f):

Using the wave speed equation v = f · λ:

f =

v

λ

=

340

0.0080

f = 42,500 Hz (or 42.5 kHz)

(ii) Calculate the Total Distance Traveled by the Pulse (S):

S = v × t = 340 × 0.10

S = 34 m

(iii) Calculate the Distance of the Prey from the Bat (d):

Since the pulse must travel to the prey and back, the distance to the prey is half of the total distance traveled by the sound wave:

d =

S

2

=

34

2

d = 17 m

Result:

(i) The frequency of the wave is 42,500 Hz.
(ii) The total distance covered by the wave is 34 m.
(iii) The prey is located at a distance of 17 m from the bat.