Q.A string 1m long is fixed at one end. Transverse vibrations of frequency 15 Hz are imposed at the free end. Due to this, a stationary wave with four complete loops, is produced on the string. Find the speed of the progressive wave which produces the stationary wave. [Hint: Remember that the free end is an antinode.]

[!TLDR] Free end = antinode, fixed end = node (like a closed pipe): with 4 complete loops plus the closing quarter-loop at the free end, $L=9\lambda/4$, so $\lambda=4L/9=4/ …

The string is fixed at one end (a NODE there) and free at the other, and the hint confirms the free end is an ANTINODE -- exactly the same boundary conditions as a pipe closed at one end (Long Answer 8), so the allowed lengths are odd multiples of a quarter-wavelength, $L=(2p+1)\lambda/4$. The pattern is described as having 'four complete loops': counting a full loop as one half-wavelength segment between successive nodes, four complete loops account for $4\times2=8$ quarter-wavelength segments, and the pattern must additionally end in a final quarter-loop reaching the antinode at the free end -- giving a total of $8+1=9$ quarter-wavelength segments, i.e. $L=\dfrac{9\lambda}{4}$ (matching $p=4 …

Recognise the fixed-free string as having closed-pipe-style boundary conditions, count the four complete (half-wavelength) loops plus the closing quarter-loop at …

Treating 'four loops' as simply $L=4\times(\lambda/2)=2\lambda$ (as if BOTH ends were nodes, the ordinary fixed-fixed string case), which misses the extra quarter-loop ne …