CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

The amplitude of a wave disturbance propagating in the positive x-direction is given by:
y=11+x2 at t=0 and y=1[1+(x1)2] at t=2 sec. Where, x and y are in meters. If the shape of the wave disturbance does not change during the propagation, what is the velocity of the wave?

A
1 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.5 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.5 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D 0.5 m/s
From the given y(x) equations at two different times we can see that the value of y is maximum (=1 m) at x=0 at t=0 and at x=1 at t=2.
So, the peak of the wave pulse has travelled a distance of 1 m (from x=0 to x=1) in 2 s along positive x - direction.
Hence, the wave velocity is:
v=12=0.5 m/s
The correct answer is Option C:0.5 m/s

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Acceleration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon