A particle executes shm on a straight line at two positions its velocity is u and v while acceleration alpha and beta respectively the displacement between these two positions is
Answers
Answered by
2
A particle is executing SHM along a straight line. Its velocities at distance x1 and x2 are v1 and v2, respectively. What is its time period?
Still have a question? Ask your own!
What is your question?
11 ANSWERS

Vinod Gohel, former Retired professor of physics at Gujarat University (1962-2003)
Updated Jan 22, 2018 · Author has 3k answers and 1.7m answer views
If A is amplitude of S.H.M. and ω is angular frequency,the velocity as function of displacement x is given by
v=ω(A2−x2)−−−−−−−−√v=ω(A2−x2).
Squaring on both sides,we have, v2=ω2(A2−x2)v2=ω2(A2−x2) .
Then, v21=ω2(A2−x21)v12=ω2(A2−x12)…………………..(1).
Similarly, v22=ω2(A2−x22)v22=ω2(A2−x22)………………….(2)
Subtracting (1) from(2), we get,
v22−v21=ω2(x21−x22)v22−v12=ω2(x12−x22). Taking ω=2pi/T=2pi/Tand making T subject of the formula,we finally get
T=2π(x21−x22)√(v22−v21)√T=2π(x12−x22)(v22−v12).
Still have a question? Ask your own!
What is your question?
11 ANSWERS

Vinod Gohel, former Retired professor of physics at Gujarat University (1962-2003)
Updated Jan 22, 2018 · Author has 3k answers and 1.7m answer views
If A is amplitude of S.H.M. and ω is angular frequency,the velocity as function of displacement x is given by
v=ω(A2−x2)−−−−−−−−√v=ω(A2−x2).
Squaring on both sides,we have, v2=ω2(A2−x2)v2=ω2(A2−x2) .
Then, v21=ω2(A2−x21)v12=ω2(A2−x12)…………………..(1).
Similarly, v22=ω2(A2−x22)v22=ω2(A2−x22)………………….(2)
Subtracting (1) from(2), we get,
v22−v21=ω2(x21−x22)v22−v12=ω2(x12−x22). Taking ω=2pi/T=2pi/Tand making T subject of the formula,we finally get
T=2π(x21−x22)√(v22−v21)√T=2π(x12−x22)(v22−v12).
Similar questions