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katrin2010 [14]
3 years ago
11

A simple pendulum oscillates back and forth with a maximum angular displacement of pi/12 radians. At t=1.2s, the pendulum is at

its maximum displacement. The length (L) of the pendulum is 35cm and the mass (m) of the bob attached to the end of it is 300g.
(a) What's the angular frequency of this oscillation?
(b) What's the angular position as a function of time?
(c) What's the angular speed as a function of time?
(d) If damping is introduced to this oscillator, would the total energy of the pendulum to increase, decrease, or stay the same?
(e) This simple pendulum is placed next to a physical pendulum that's a rod with a mass of m = 300g and length L =35cm. Is the physical pendulum’s angular frequency larger, smaller, or the same as the simple pendulum?
Physics
1 answer:
Zielflug [23.3K]3 years ago
6 0

Answer:

a)  w = 1.08 rad / s , b) θ = π / 12 cos (1.08 t - 1,296) , c)  w = - 0.2827 sin (1.08t - 1,296) , d) energy decrease

Explanation:

a) The oscillatory movement of a simple pendulum has the angular velocity

      w = √(L / g)

      w = √(0.35 / 0.300)

      w = 1.08 rad / s

b) it is requested to find the equation of angular displacement

       θ = θ₀ cos (wt + φ)

We replace

       θ = π / 12 cos (1.08 t + φ)

To find the constant let's use the value they give

       θ = θ₀     for   t = 1.2 s

       θ = θ₀ cos 1.08 1.2 + φ)

       Cos (1,296 + φ) = 1

        1,296 + φ = cos⁻¹ 1  

Remember that the angles must be in radians

         φ = 0 - 1,296

The final equation is

        θ = π / 12 cos (1.08 t - 1,296)

c) the angular velocity is

         w = dθ / dt

         w = π / 12 (- 1.08 sin (1.08t - 1,296))

         w = - 0.2827 sin (1.08t - 1,296)

d) if addamping force is included, part of the energy dissipates, therefore the total energy must decrease

e) The period of a physical pendulum is

      w = √ (mg d / I)

Where I is the moment of inertia that is of the form

      I = cte m R²

     w = √ (m g d / cte md2) = √ (g / d)  √(1/cte)

Simple pendulum wo = √ (g / d)

       w = w₀  1/√cont

As we see the angular velocity of this pendulum change due to the constant that accompanies the moment of inertia. In general this constant is less than by which the angular velocity the angular velocity increases

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I = 0.287 MR²

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\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

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