Question

In the following properties of a wave, the one which is independent of the other is
(A) Amplitude
(B) Velocity
(C) Wavelength
(D) Frequency

Answer 1

Hint To answer this question, we need to try to identify the relationship between each of the quantities given in the question. Then we need to obtain a relation mathematically, which will relate the three of the four quantities given. The fourth quantity, which will not be able to fit in that mathematical relation, will be the final answer.

 Formula Used: The formulae used in this solution are
 $\Rightarrow v = \dfrac{\lambda }{T} $
 $\Rightarrow \nu = \dfrac{1}{T} $
Here $ v $ is the velocity of a wave having the wavelength $ \lambda $ , the frequency $ \nu $ , and the time period $ T $ .

Complete step by step answer
To start this question, we begin finding the quantity from the options, whose mathematical formula is known to us.
As we know the formula for the velocity of a wave, so we choose it as our first quantity.
We know that the velocity of a wave is related to its wavelength by
 $\Rightarrow v = \dfrac{\lambda }{T} $ ……………………….(1)
So, we have related two of the four quantities given in the options. Now we need to search for the third quantity.
As the frequency is related to the time period $ T $ of the wave, so we choose it as our third quantity.
We know that the frequency is given by
 $\Rightarrow \nu = \dfrac{1}{T} $ ……………………..(2)
Putting (2) in (1) we get
 $\Rightarrow v = \lambda \nu $
So, the velocity of a wave is equal to the product of its wavelength and frequency.
Now, the above relation is valid for any given amplitude of the wave. So, this means that none of the other quantities can decide its value. Thus it cannot be related to them by any relation.
Hence, the correct answer is option A.

Note
You might be wondering why the amplitude of a wave is independent. This is due to the fact that the amplitude of a wave is a function of the initial energy only, which is given to it. The energy is provided externally to the particles of the medium through which the wave propagates.