The time between the successive crusts will give the time period of the wave.
So, the time period, T of the given wave is 0.2 s.
The frequency of a wave is the reciprocal of its time period. That is,
Frequency, ν =
=
= 5 Hz
Thus, the frequency of the given travelling periodic wave is 5 Hz.
a) The speed of the wave will increase by a factor .
b) The speed of the wave will halve
c) The speed of the wave will double
Explanation:
a)
The speed of a standing wave on a string is given by
where
T is the tension in the string
m is the mass of the string
L is the length of the string
In this part of the problem, the tension in the string is doubled, so that the new tension is
T' = 2T
Substituting into the equation, we find the new speed of the wave in the string:
So, the speed will increase by a factor .
b)
We can solve also this part by referring to the formula
where
T is the tension
m is the mass
L is the length
In this case, the string mass is quadrupled, so the new mass is:
m' = 4m
Substituting into the equation, we find what happens to the speed of the wave:
So, the speed of the wave will halve.
c)
Again, we can solve this part by referring to the same equation
where
T is the tension
m is the mass
L is the length
In this case, the length of the string is quadrupled, so the new length is:
L' = 4L
Substituting into the equation, we find that the new speed is:
So, the speed of the wave will double.
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Answer:
None of the wavelength is in the visible range
Explanation:
Constructive interference of the reflected waves for different wavelengths can be estimated using:
λ = 2nd/m
where m is 1,2,3, ...
Therefore:
m=1, λ = 750 nm
m=2, λ = 750/2 = 375 nm
The limits of eye's sensitivity is between 430 nm and 690 nm. Beyond this range, the eye's sensitivity drops to approximately 1% of its maximum value.