Question

two pendula are released from rest at the same height h above their lowest points. how do the maximum velocities of the two pendulums compare? note that pendulum 2 has a string with length twice that of pendulum 1.

Answer 1

The simple pendulum concept and rotational kinematics relationships allow finding the result for the velocity relationship

•  The shortest pendulum (w₁) has more speed than the longest pendulum (w₂)

               w₁ = $\sqrt{2}$  w₂

 

The simple harmonic movement is a periodic movement where the restoring force is proportional to the elongation, in the case of the simple pendulum it is fulfilled for angles less than 10º.  The expression for motion is

          θ  = θ₀ cos ωt

         ω² = g / l

the angular velocity is defined by

         w = $\frac{d \theta}{dt}$  

         w = - θ₀ ω sin ωt

Where θ₀ is the initial angle or amplitude, w is the angular velocity, t the time, g the acceleration of gravity and l the length of the pendulum

Let's find the angular velocity for each pendulum

          ω ₁ = $\sqrt{ \frac{g}{l_1} }$

Indicate that the second pendulum length is

          l₂ = 2 l₁

         ω₂ = $\sqrt{\frac{g}{2l_1} }$  

the pendulum speed is maximum at the lowest point of the trajectory that occurs when the sine function has its maximum value

          w =  θ₀ω  

Suppose that the two pendulums are released with the same initial angle

         

Pendulum 1

          w₁ =  θ₀ $\sqrt{\frac{g}{l_1} }$

Pendulum 2

          w₂ = θ₀  $\sqrt{\frac{g}{2l_1} }$

We look for the relationship between the two speeds

           $\frac{w_1}{w_2} = \sqrt{2}$  

In conclusion, using the simple pendulum concept and the rotational kinematics relations we can find the result for the velocity relation

 

• The shortest pendulum (w₁) has more speed than the longest pendulum (w₂)

             w₁ = √2 w₂

Learn more here:  brainly.com/question/24159297