Question

We consider air as a fluid. The density at 15.0°C is 1.225 mg/cm3 , and the bulk modulus is 1.416×105 Pa. In these conditions, a bat flies toward a wall, emitting a steady sound of frequency 1.70 kHz. This bat hears its own sound plus the sound reflected by the wall. a) Calculate the speed of sound in air at 15.0°C? b) How fast should the bat fly in order to hear a beat f

Answer 1

The complete b part is;

b) How fast should the bat fly in order to hear a beat frequency of 8.00 Hz?

Answer:

A) V = 340 m/s

B) v = 0.78 m/s

Step-by-step explanation:

A) We are given;

Bulk modulus; B = 1.416 × 10^(5) Pa = 1.416 × 10^(5) N/m²

Density; ρ = 1.225 mg/cm³ = 1.225 kg/m³

Formula for speed of sound in air in a fluid is;

V = √(B/ρ)

V = √(1.416 × 10^(5)/1.225)

V = 340 m/s

B) We are given frequency of bat; f_o = 1.7 KHz = 1700 Hz

Since we know the speed of sound.

Thus, incident frequency is;

f_incid = [(340 - 0)/(340 - v)] × f_o

f_incid = [340/(340 - v)] × f_o

Now, the wall is stationary. Thus, the reflected frequency will be equal to the incident frequency.

f_incid = f_refl

Now, new frequency heard by the bat is given by;

f_new = [(340 + v)/340] × f_refl

Since f_incid = f_refl and f_incid = [(340 - 0)/(340 - v)] × f_o

Thus;

f_new = [(340 + v)/340] × [(340/(340 - v)] × f_o

This gives;

f_new = [(340 + v)/(340 - v)] × f_o

We are given beat frequency as 8 Hz. Thus;

f_new - f_o = 8 Hz

f_o = 1700;

[(340 + v)/(340 - v)] × 1700] - 1700 = 8

[(340 + v)/(340 - v)] = 1708/1700

[(340 + v)/(340 - v)] = 1.00471

340 + v = 1.00471(340 - v)

340 + v = 341.6014 - 1.00471v

v + 1.00471v = 341.6014 - 340

2.00471v = 1.6014

v = 1.6014/2.00471

v = 0.78 m/s