If the displacement x and velocity v of a particle executing S.H.M are related through the expression 4v2=25−x2, then its maximum displacement in meters is
A
1
B
2
C
5
D
6
Medium
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Updated on : 2022-09-05
Solution
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Correct option is C)
v=ωA2−x2
4v2=52−x2⟹v=21(52−x2) Comparing the equation with the standard equation we get A=5m