a sound wave of wavelength 0.33 m has a time period of 10^-3s.if the time period is decreased by 10^-4 , calculated the wavelength of the new wave
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Answered by
2
Hey There!!!
For this question, we are going to use the following concepts:
Here, the sound is always travelling in the same medium. So, speed of the sound wave is going to be constant.
So,
Here, we can take the values as:
Using the relation we have:
So, we can write a relation
Thus, the new wavelength is 0.033 metres
Hope it helps
Purva
Brainly Community
For this question, we are going to use the following concepts:
Here, the sound is always travelling in the same medium. So, speed of the sound wave is going to be constant.
So,
Here, we can take the values as:
Using the relation we have:
So, we can write a relation
Thus, the new wavelength is 0.033 metres
Hope it helps
Purva
Brainly Community
QGP:
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Answered by
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time period = T = 10^-3s
wavelength = a = 33×10^-2 m
frequency = f = ?
velocity = v = ?
T = 1/f
fT = 1
f = 1/T
f = 1/10^-3
f = 10³Hz
so frequency of old wave is 10³Hz
then frequency of new wave is 10⁴Hz
velocity= wavelength×frequency
v = an
if one wave time period decrease then it's frequency increase means it's speed constant
af= a¹f¹
33×10^-2× 10³ = 10⁴×a¹
33×10¹/10⁴ = a¹
33×10^-3 = a¹
so wavelength of new wave is 33×10^-3m
wavelength = a = 33×10^-2 m
frequency = f = ?
velocity = v = ?
T = 1/f
fT = 1
f = 1/T
f = 1/10^-3
f = 10³Hz
so frequency of old wave is 10³Hz
then frequency of new wave is 10⁴Hz
velocity= wavelength×frequency
v = an
if one wave time period decrease then it's frequency increase means it's speed constant
af= a¹f¹
33×10^-2× 10³ = 10⁴×a¹
33×10¹/10⁴ = a¹
33×10^-3 = a¹
so wavelength of new wave is 33×10^-3m
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