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How is the period of a pendulum affected when its point of suspension is (a) moved horizontally in the plane of oscillation with acceleration a; (b) moved vertically upward with acceleration a; (c) moved vertically downward with acceleration a g? Which case, if any, applies to a pendulum mounted on a cart rolling down an inclined plane?

How is the period of a pendulum affected when its point of suspension is (a) moved horizontally in the plane of oscillation with acceleration a; (b) moved vertically upward with acceleration a; (c) moved vertically downward with acceleration a g? Which case, if any, applies to a pendulum mounted on a cart rolling down an inclined plane?

Grade:upto college level

2 Answers

Jitender Pal
askIITians Faculty 365 Points
8 years ago
The period T of a simple pendulum is defined as,
T = 2π√L/g
Here L is the length of the pendulum and g is the free fall acceleration.
So, the period of the simple pendulum is independent of the mass m of the suspended particle.
(a)
The period T of the pendulum, when its point of suspension is moved horizontally in the plane of oscillation with acceleration a will be,
T = 2π√L/g.
The pendulum will oscillate horizontally in the plane. So, it will never change, when its point of suspension is moved horizontally
(b)
The period T of the pendulum, when its point of suspension is moved vertically upward with acceleration a will be,
T = 2π√L/a-g.
(c)
The period T of the pendulum, when its point of suspension is moved vertically downward with acceleration a < g will be,
T = 2π√L/g-a
But when, a > g, the time period of the pendulum will be imaginary. So, there is no oscillation in this case.
The period of oscillation, if any, applies to a pendulum mounted on a cart rolling down an inclined plane will be,
T = 2π√L/g cosθ-a
Here, θ is the angle of inclination
pa1
357 Points
8 years ago
 
The period T of a simple pendulum is defined as, T = 2π√L/g Here L is the length of the pendulum and g is the free fall acceleration. So, the period of the simple pendulum is...

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