$A$ stretched wire $0.5\, m$ long is made to vibrate in two different modes as shown in diagrams $(A)$ and $(B)$ given below:
$(i)$ If the wavelength of the wave produced in mode $(A)$ is $1\, m$,what is the wavelength of the wave produced in mode $(B)$ of the following diagram?
$(ii)$ In which case is the note produced louder? Give a reason for your answer.
$(iii)$ In which case is the pitch of the note produced higher? Give a reason for your answer.

(N/A) $(i)$ In mode $(A)$,the wire vibrates in one loop,so the length $L = \lambda/2$,which means $\lambda = 2L = 2 \times 0.5\, m = 1\, m$. In mode $(B)$,the wire vibrates in two loops,so the length $L = \lambda'$. Thus,$\lambda' = L = 0.5\, m$. Therefore,the wavelength in mode $(B)$ is $0.5\, m$.
$(ii)$ The note produced in mode $(A)$ is louder. This is because the amplitude (maximum displacement from the mean position) of the wave in mode $(A)$ is greater than that in mode $(B)$. Since loudness is directly proportional to the square of the amplitude,the sound in mode $(A)$ is louder.
$(iii)$ The pitch of the note produced in mode $(B)$ is higher. This is because pitch is directly related to the frequency of the sound wave. As seen from the diagram,mode $(B)$ has two loops,meaning it vibrates at a higher frequency compared to mode $(A)$,which has only one loop. Higher frequency results in a higher pitch.