Answer:
Hence, The speed of transverse waves in the wire is 327m/s
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
Answer:
The speed of transverse waves in the wire is 326.95m/s
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 = [tex]\lambda/4[/tex]
So,
[tex]\lambda = 20cm \times 4 \\\\= 80cm \\\\= 0.8m[/tex]
Frequency = [tex]\frac{c}{\lambda}[/tex]
[tex]= \frac{340}{0.8} \\\\=425Hz[/tex]
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
