The particle executing SHM in a straight line has velocities 8 m/s p 7 m/s and 4 m/s at three points having distance 1 m from each other. what will be maximum velocity of particle ?
Answers
The maximum velocity of the particle will be √65 m/s
Explanation :
The maximum velocity of a particle executing SHM is given by,
Vmax = Aw
The velocity at any point x is given by,
V² = w²(A²-x²)
where w = angular velocity,
A = amplitude
x = distance from central position
let the three points A,B,C are at a distance of d, d+1, d+2 meters from the central position,
hence,
V₁² = w²(A²-d²) = 64.............eq1
V₂² = w²(A²-(d+1)²) = 49.............eq2
V₃² = w²(A²-(d+2)²) = 16............eqn3
subtracting eqn2 from eqn1 we get
15 = w²(2d+1).................eqn4
subtracting eqn3 from eqn2 we get
33 = w²(2d+3)...........eqn5
dividing eqn5 and eqn4 we get
(2d+3)/(2d+1) = 33/15
=> 30d + 45 = 66d + 33
=> d = 1/3
putting the value of d in eqn4 we get,
15 = w²(2/3 + 1 )
=> w = 3
putting the value of w and d in eqn1 we get
w²(A²-d²) = 64
=> 9(A²-1/9) = 64
=> A² = 64/9 + 1/9 = 65/9
=> A = √65/3
Hence maximum velocity of the particle,
Vmax = Aw = √65
Answer:
THE ANSWER OF THE ABOVE QUESTION IS √65