a simple pendulum has a period of 2.0s and magnitude of swing 5cm.find i-the velocity of the bob ii-the acceleration of the bob
See example 3.
https://books.google.com/books?id=BukDAAAAMBAJ&pg=PA1433&lpg=PA1433&dq=a+simple+pendulum+has+a+period+of+2.0s+and+magnitude+of+swing+5cm.find+i-the+velocity+of+the+bob+ii-the+acceleration+of+the+bob&source=bl&ots=gDxmNuClKx&sig=agV4IL0lNHcEuDWK8urzpBYLHkA&hl=en&sa=X&ved=0ahUKEwiypPmekYbUAhVF6YMKHe7lDE4Q6AEILjAB#v=onepage&q&f=false
To find the velocity and acceleration of a simple pendulum with a given period and magnitude of swing, we can use the equations of motion for simple harmonic motion.
First, let's find the angular frequency (ω) of the pendulum. The angular frequency is given by the formula ω = 2π / T, where T is the period of the pendulum.
Given that the period of the pendulum is 2.0 seconds, we can calculate the angular frequency:
ω = 2π / 2.0 = π rad/s
Now, let's calculate the velocity and acceleration of the pendulum bob.
(i) Velocity of the bob:
The velocity of the bob at any point in its swing is given by the formula v = ωA, where A is the amplitude of the swing. In this case, the amplitude is equal to the magnitude of the swing, which is 5 cm or 0.05 m.
v = ωA = (π rad/s)(0.05 m) = 0.157 m/s
Therefore, the velocity of the pendulum bob is 0.157 m/s.
(ii) Acceleration of the bob:
The acceleration of the bob at any point in its swing is given by the formula a = -ω²x, where x is the displacement from the equilibrium position. For a simple pendulum, the maximum displacement is equal to the amplitude.
a = -ω²A = -(π rad/s)²(0.05 m) = -0.247 m/s²
Therefore, the acceleration of the pendulum bob is -0.247 m/s². Note that the negative sign indicates that the acceleration is directed towards the equilibrium position.
In summary:
(i) The velocity of the bob is 0.157 m/s.
(ii) The acceleration of the bob is -0.247 m/s².