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Question: Transverse wave of amplitude 5 cm is generated at one end (x=0) of a long string by a tuning fork of...

Transverse wave of amplitude 5 cm is generated at one end (x=0) of a long string by a tuning fork of frequency 500 Hz. At a certain instant of time, the displacement of a particle A at x=100 cm is -5 cm and of particle B at x=200 cm is +5 cm. What could be the wavelength of the wave?

A. 2m2m
B. 3m3m
C. 4m4m
D. 5m5m

Explanation

Solution

There are two kinds of waves normally. One will be a transverse wave and the other will be a longitudinal wave. In case of transverse waves the direction of propagation of waves is perpendicular to the direction of vibration of the particles of the medium. Transverse wave has crests and troughs.

Complete step by step answer:
In case of the longitudinal wave particles of the medium of propagation vibrates in the direction of propagation of the wave. This longitudinal wave has compressions and rarefactions. In case of longitudinal waves particles vibrate in to and fro motion. Transverse waves can propagate without medium too whereas longitudinal waves require medium to propagate. Example for transverse wave is light whereas example for the longitudinal wave is sound.
The given wave is the transverse wave and amplitude is given as 5cm.
Displacement of particles at
x=100cm=1mx = 100cm = 1m is 5cm - 5cm
x=200cm=2mx = 200cm = 2m is +5cm + 5cm
Now coming to crests and troughs in case of transverse wave, crest is a point where the displacement of the particle will be the positive maximum i.e positive amplitude and trough is the point where displacement of the particle is negative maximum I.e negative amplitude.
So particle A is at the trough and particle B is at the crest.
Distance between the adjacent trough and crest will be λ2\dfrac{\lambda }{2} where λ\lambda is the wavelength of the wave.
It is given that the distance between trough and crest is 200100=100cm=1m200 - 100 = 100cm = 1m
Which means
\eqalign{ & \dfrac{\lambda }{2} = 1m \cr & \Rightarrow \lambda = 2m \cr}
Hence the wavelength of the wave will be 2m
Hence option A will be correct.

Note:
Actually the frequency given over in this particular problem is to confuse us. It is not required to solve this problem. There might be the case that particles A and B are not adjacent crest and troughs. But since nothing is given we assume to be adjacent and solve for the wavelength.