Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A transverse wave along a string is given by y=2sin2π3t-x+π4, where, x and y are in cm and t in second. Find the acceleration of a particle located at x=4 cm at t=1 s.
PhysicsWaves and SoundNEET
Options:
  • A 362 π2 cm s-2
  • B 36π2 cm s-2
  • C -362 π2 cm s-2
  • D -36π2 cm s-2
Solution:
2944 Upvotes Verified Answer
The correct answer is: -362 π2 cm s-2
We know that v=dydt

=ddt 2sin2π3t-x+π4

a=dvdt=ddt12πcos2π3t-x+π4

a=-72π2sin2π3t-x+π4

at t=1 and x=4 cm

a=-36 2π2 cm s-2

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.