For a transverse wave travelling along a straight line, the distance between two peaks (crests) is 5 m, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in m) of the are :

Solution

Physics sir · 3 steps

Crest to crest is a whole number of wavelengths; crest to trough is an odd number of half wavelengths.

1. Two conditions. The two crests need not be neighbours, so 5 m can be any whole number n of wavelengths. A crest and a trough are half a wavelength apart, plus any whole number of wavelengths, so 1.5 m must be an odd number of half wavelengths.

   5 = nλ, 1.5 = (2m+1)λ/2

2. Combine them. Dividing gives 10/3 = 2n/(2m+1), so 3n = 5(2m+1). Then 2m+1 must be an odd multiple of 3, that is 2m+1 = 3(2k+1). This gives λ = 3/(2m+1) = 1/(2k+1).

   λ (m)	5/λ (whole number?)	1.5/λ (0.5, 1.5, 2.5, ...?)	Allowed?
   1	5, yes	1.5, yes	yes
   1/3	15, yes	4.5, yes	yes
   3	5/3, no	0.5, yes	no
   2	2.5, no	0.75, no	no
   1/2	10, yes	3, no	no

3. So. The allowed wavelengths are 1, 1/3, 1/5, … m, which is option (b).

The idea to keep. Same phase means separated by nλ. Opposite phase means separated by (2m+1)λ/2. When both distances are given, the wavelength has to satisfy both conditions at once.

Answer from the NTA final answer key. Solution by PrepWiser.