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This is second of our many quizzes on Pendulum energy. Click on this to access more such competitive quizzes on Mechanics.

Pendulum energy Quiz 2

physics

Will the mechanical energy in a damped pendulum exhibiting damped harmonic oscillation be, 

A damped pendulum is one in which the motion of the bob is restricted or has friction such as the air resistance. Due to this there is a loss in mechanical energy in the form of heat or sound and mechanical energy is not conserved.

  A damped pendulum is one in which the motion of the bob exhibiting damped harmonic oscillation reduces due to external force or friction such as the air resistance. Due to this there is a loss in mechanical energy as the bob oscillates in the form of heat or sound. Hence the total mechanical energy in a damped pendulum is not conserved.

Consider a yo-yo of mass $12$ g and the length of its thread is $14$ cm, it is executing simple harmonic motion. If at a certain time the thread makes an angle of ${{40}^{0}}$ with the normal, then what is the potential energy at this moment? 

The gravitational potential energy is given by, $P.E.=mgl(1-\cos \theta )$.Here, $m$ is mass, $g$ is the acceleration due to gravity, $l$ is the length of the pendulum and $\theta $ is the angle of displacement. Plug in the values given in the question to the potential energy.

  $ \text{The formula for gravitational potential energy is given by,}\\ P.E.=mgl(1-\cos \theta )….(i)\\ \text{ According to the question,}\\ m=12\,g=0.012\,kg, l=14\,cm=0.14\,m\,\text{and}\, \theta ={{40}^{0}}\\ \text{ Putting these values in equation (i), we get,}\\ P.E.=0.012\times 10\times 0.14\times (1-\cos {{40}^{0}})\\ \therefore P.E.=3.9\times {{10}^{3}}\,J$ 

If a simple pendulum of mass is $36$ g, the length is $38$ cm, the velocity at an angle of ${{35}^{0}}$ is $5.8$ m/s. What is the total mechanical energy at this point? $(\text{Take g=10 m/s})$ 

The mechanical energy M.E. is always conserved, given by, $M.E.=\dfrac{1}{2}m{{v}^{2}}+mgl(1-\cos \theta )$.Here, $m$ is mass, $g$ is the acceleration due to gravity, $l$ is the length of the pendulum and $\theta $ is the angle of displacement and $v$ is the velocity. Plug in the known values and find the total mechanical energy.

 $\text{ The total mechanical energy in a simple pendulum is}\\ \text{the combination of kinetic energy and gravitational potential energy.}\\ M.E.=K.E.+P.E.….(i)\\ \text{The formula for kinetic energy and potential energy is given by,}\\ K.E.=\dfrac{1}{2}m{{v}^{2}},P.E.=mgl(1-\cos \theta )\\ M.E.=\dfrac{1}{2}m{{v}^{2}}+mgl(1-\cos \theta )….(ii)\\ \text{ According to the question, mass is m=36 g=0.036 kg,}\\ \text{length is l}=38\,cm=0.38\,m, \theta ={{35}^{0}}\,\text{and}\,v=5.8\,m/s.\\ \text{ Putting the above values in equation (ii), we get,}\\ M.E.=\dfrac{1}{2}\times 0.036\times {{(5.8)}^{2}}+0.036\times 10\times 0.38\times (1-\cos {{35}^{0}})\\ \therefore M.{{E}_{2}}=0.63\,J$

Consider a pendulum of mass is $86$ g, the length is $21$ cm, if the velocity at the initial angular displacement of ${{45}^{0}}$ is $5.9$ m/s, then what is the potential energy at this angle?$(\text{Take g=10 m/s})$ 

The gravitational potential energy is given by, $P.E.=mgl(1-\cos \theta )$.Here, $m$ is mass, $g$ is the acceleration due to gravity, $l$ is the length of the pendulum and $\theta$ is the angle of displacement. Plug in the values given in the question to the potential energy at an angle $\theta ={{45}^{0}}$. 

 $\text{ The formula for gravitational potential energy is given by,}\\ P.E.=mgl(1-\cos \theta )….(i)\\ \text{ According to the question,}\\ m=86 g=0.086 kg, l=21cm=0.21\,m\,\text{and}\, \theta ={{45}^{0}}\\ \text{ Putting these values in equation (i), we get,}\\ P.E.=0.086\times 10\times 0.21\times (1-\cos {{45}^{0}})\\ \therefore P.E.=52.9\,mJ$ 

A pendulum of mass is $90$ g and length is $38$ cm, is in S.H.M. What is the angle at which the K.E. and P.E. of the pendulum be equal to each other, if the velocity at this point is $12$ m/s? 

Here, the kinetic energy is equal to the potential energy, so kinetic energy and potential energy is given by, $\dfrac{1}{2}m{{v}^{2}}=mgl(1-\cos \theta )$.Here, $m$ is mass, $v$ is velocity, $g$ is the acceleration due to gravity, $l$ is the length of the pendulum and $\theta $ is the angle of displacement.Plug in the values given in the question to find the unknown angle $\theta $.

 $\text{ The formula for kinetic energy and potential energy is given by,}\\ K.E.=\dfrac{1}{2}m{{v}^{2}},P.E.=mgl(1-\cos \theta )….(i)\\ \text{ According to the question,}\\ m=90g=0.090kg,l=38cm=0.38m\,\text{and}\,v=12\,m/s\\ \text{Here, the kinetic energy is equal to the potential energy,}\\ \text{so putting these values in equation (i) and equating, we get,}\\ K.E.=P.E.\\ \Rightarrow \dfrac{1}{2}\times 0.090\times {{(12)}^{2}}=0.090\times 10\times 0.38\times (1-\cos \theta )\\ \therefore \theta ={{36}^{0}}$ 