Question

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column open at one end is gradually increased from 0 until the 1st position of resonance is observed at 20 cm. The wire is 115.4 cm long and is vibrating in its third harmonic.

Find the speed of transverse waves in the wire if the speed of sound in air is 340 m/s. Answer in units of m/s.

Answer 1

The speed of transverse waves in the wire is 326.95m/s

Step-by-step explanation:

Find frequency of sound in pipe

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance

At 1st resonance (fundamental) pipe length =
`\lambda/4`

So,

`\lambda = 20cm * 4 \\\\= 80cm \\\\= 0.8m`

Frequency =
`(c)/(\lambda)`

`= (340)/(0.8) \\\\=425Hz`

The wire (which we now know is vibrating at 425Hz)

For the wire, there are 3 segments, so 4 nodes (nodes at each end and 2 in middle).

Node-node distance = 115.4 / 3

= 38.47cm

wires λ / 2 = 38.47cm (as non-to-node is half a wavelength)

So wires λ = 38.47 x 2

= 76.93cm

Speed on wire = wires λ x freq

= 0.7693m x 425Hz

= 326.95m/s

Hence, The speed of transverse waves in the wire is 326.95m/s

Answer 2

Hence, The speed of transverse waves in the wire is 327m/s

Step-by-step explanation:

Since one end of the pipe is opened, the length of the pipe can be written as

L = ¼λ

λ = 4L

Given that, at first resonance, length is

L = 20cm

Then,

λ = 4L = 4 × 20

λ = 80cm

λ = 0.8m

The wavelength of the sound wave is 0.8m

The frequency of the sound wave can be determine using wave equation

v = f λ

f = v / λ

Given that, speed of sound in air is 340m/s

f = 340/0.8

f = 425 Hz

When, the wire is in third harmonic,

The length of the pipe is given as

L = 3λ/2

2L = 3λ

λ = ⅔L

Since the length of the wire is

L = 115.4cm = 1.154m

Therefore,

λ = ⅔L

λ = 2/3 ×1.154

λ = 0.7693m

Then, using wave equation,

v = f × λ

v = 425 × 0.7693

v = 326.97 m/s

v ≈ 327m/s