wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A standing wave exists in a string of length 150 cm which is fixed at both ends with rigid supports. The displacement amplitude of a point at a distance of 10 cm from one of the ends is 53 mm. The distance between the two nearest points, within the same loop and having displacement amplitude equal to 53 mm is 10 cm. The maximum displacement amplitude of the particles in the string is


A

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

10 mm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

15 mm

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

20 mm

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

10 mm


y=A sin kx cosωt

53=A sin k×10

53=A sin k×20

After solving k=π30

53=Asin(π30×10)A=10 mm


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Building a Wave
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon