Given : L = 1 m, n = 15 Hz.
The string is fixed only at one end. Hence, an antinode will be formed at the free end. Thus, with four and half loops on the string, the length of the string is
L = \(\frac{λ}{4}+4\frac{λ}{2}=\frac{9}{4}λ\)
∴ λ = \(\frac{4L}{9}\) =\(\frac{4}{9}×1\)
v = n λ
∴ Speed of the progressive wave
v =15 x\(\frac{4}{9}\)= 6.667 m/s