Question

I will be happy if you show me your best explanation. I will provide a thumb up if you solve correctly.

8. Two identical piano wires have a fundamental frequency of 650 Hz when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of 8 beats/s when both wires vibrate simultaneously? Answer: ( which one: 0.0248 or 0.0246 and why)

9. A stone is dropped into a well. The sound of the splash is heard 3.5 s later. What is the depth of the well? Take the speed of sound in air to be 343 m/s. Answer: a=7.4 h=54.8m

Answer 1

here,

38)

The beat frequency: df = 8 Hz

If we increase string tension, then, the frequency of the sound on the
string will be higher. thus, f' = f + df = 650 + 8 = 658 Hz

The speed of a traveling sound wave on a string is given by:

v = sqrt(T/u)

With the tension on the 2nd string increased, v' = sqrt(T'/u)

where, tension T' = T + dT.

If we take the ratio of wave speeds on the strings:

v' / v = sqrt(T'/u) / sqrt(T/u) = sqrt(T'/T )here

Squaring both sides: (v'/v)^2 = (T + dT) / T = 1 + dT/T

dT/T = (v' / v)^2 - 1

But, the velocity is directly proportional to the frequency, since, v = f lamda.

So that without loss of generality, we can rewrite the tension ratio above as:

dT/T = (f' / f)^2 - 1 = (658/650)^2 - 1 = 1.0248 - 1 = 0.0248

9)

let the depth of well be h

time taken by sound , t1 = h/343 s

and

the time taken by stone to hit the water , t2 = 3.5 - t1

t2 = (3.5 - h/343 ) s

for stone

h = 0 + 0.5 * g * t2^2

h = 0 + 0.5 * 9.81 *(3.5 - h/343)^2

solving for h

h = 54.8 m

the depth of well is 54.8 m