A simple pendulum oscillates in a vertical plane. When it passes throu...
T=3mg (given)
as we know that T=3mg(1+cosA)
so,
3mg=3mg(1+cosA)
1=1+cosA
0=cosA
cos90=cosA
A=90
Hence,maximum displacement of the string of the pendulum wit respect to the vertical will be at an angle of 90 degree
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A simple pendulum oscillates in a vertical plane. When it passes throu...
We can use the conservation of energy to solve this problem. At the highest point of the oscillation, the pendulum bob has the maximum potential energy and zero kinetic energy. At the lowest point, it has the maximum kinetic energy and zero potential energy. The total energy of the system remains constant.
Let's take the zero of potential energy at the mean position of the pendulum. Then, at the highest point, the potential energy is equal to the initial potential energy plus the work done by the tension in the string. Similarly, at the lowest point, the kinetic energy is equal to the initial kinetic energy minus the work done by the tension in the string.
Let's denote the weight of the pendulum bob as W, the tension in the string as T, the length of the string as L, and the maximum angular displacement of the pendulum from the vertical as θ.
At the mean position, the tension in the string is 3W, so the initial potential energy of the pendulum is 3WL. At the highest point, the tension in the string is equal to the weight of the pendulum bob, so the potential energy is equal to (W - T)L = (W - W/3)L = 2WL/3. Thus, the work done by the tension in the string is equal to (3/3 - 2/3)WL = WL/3.
At the lowest point, the tension in the string is again equal to the weight of the pendulum bob, so the kinetic energy is equal to (1/2)mv^2 = (1/2)WL^2θ^2, where m is the mass of the pendulum bob and v is its velocity at the lowest point.
By the conservation of energy, we have:
initial potential energy = final potential energy + final kinetic energy
3WL = (2WL/3) + (1/2)WL^2θ^2
Simplifying and solving for θ, we get:
θ = sqrt(6/5) ≈ 1.095 radians ≈ 62.8 degrees.
Therefore, the maximum angular displacement of the pendulum from the vertical is approximately 62.8 degrees.
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