The basic relationship between frequency, wavelength and speed of a wave is
[tex]\lambda= \frac{v}{f} [/tex] (1)
where
[tex]\lambda[/tex] is the wavelength
v is the wave speed
f is the frequency
The problem says the two waves have same speed, so [tex]v_1 = v_2[/tex], and that the first wave has twice the frequency of the second wave, so
[tex]f_1 = 2 f_2[/tex]
If we use eq,(1), we can compare the two wavelengths
[tex]\lambda_1 = \frac{v_1}{f_1}= \frac{v_2}{2f_2}= \frac{1}{2} \frac{v_2}{f_2} = \frac{1}{2} \lambda_2 [/tex]
Where in the last step we used [tex]\lambda_2 = \frac{v_2}{f_2} [/tex]. Therefore, the first wave has half the wavelength of the second wave, so the correct option is option 2).
