Answer:
0.75m
Explanation:
Let wavelength = dThen length between two successive antinodes will be d/2.
frequency = 32Hz
velocity = 48m/s
v = fd
d = v/f
d = 48/32 = 3/2 m = wavelength
length between two successive antinodes = d/2 = 3/4 = 0.75m
First of all, let's calculate the wavelength of the stationary wave :
[tex]v = f \times \lambda[/tex]
[tex] \implies \: \lambda = \dfrac{v}{f} [/tex]
[tex] \implies \: \lambda = \dfrac{48}{32} [/tex]
[tex] \implies \: \lambda = 1.5 \: metre[/tex]
So, let the distance be d :
[tex] \implies \: d = \dfrac{ \lambda}{2} [/tex]
[tex] \implies \: d = \dfrac{1.5}{2} [/tex]
[tex] \implies \: d = 0.75 \: m[/tex]
So, is between successive antinode is 0.75 metres.
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Answers & Comments
Answer:
0.75m
Explanation:
Let wavelength = d
Then length between two successive antinodes will be d/2.
frequency = 32Hz
velocity = 48m/s
v = fd
d = v/f
d = 48/32 = 3/2 m = wavelength
length between two successive antinodes = d/2 = 3/4 = 0.75m
DISTANCE BETWEEN SUCCESSIVE ANTI NODES IS 0.75 METRES.
Given:
To find:
Calculation:
First of all, let's calculate the wavelength of the stationary wave :
[tex]v = f \times \lambda[/tex]
[tex] \implies \: \lambda = \dfrac{v}{f} [/tex]
[tex] \implies \: \lambda = \dfrac{48}{32} [/tex]
[tex] \implies \: \lambda = 1.5 \: metre[/tex]
So, let the distance be d :
[tex] \implies \: d = \dfrac{ \lambda}{2} [/tex]
[tex] \implies \: d = \dfrac{1.5}{2} [/tex]
[tex] \implies \: d = 0.75 \: m[/tex]
So, is between successive antinode is 0.75 metres.