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Arturiano [62]
3 years ago
7

The pendulum with the greatest frequency is one with the

Physics
1 answer:
Jlenok [28]3 years ago
7 0

Answer:

The pendulum with the greatest frequency is one with short length.

Explanation:

A simple pendulum is a device which consists of mass m hanging from the string of length L attached to the some point.When displaced and released its swings back and forth with periodic motion.

The frequency of simple pendulum is given by

    f = \frac{1}{T}

where T is the time period.    

f =1/2π√g/l  

where , l is the length of pendulum

             g is acceleration due to gravity.

The frequency of simple pendulum is defined as how much time or the distance covered by the bob in one second

From the given equation we can find that the frequency of the simple pendulum is<em> inversely proportional</em> to square root of length.Hence the frequency will be maximum if the length of the bob is short.

 

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Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom
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Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

   final velocity of Tarzan = v_f

law of conservation of energy

  PE_i + KE_i = PE_f + KE_f

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i = \dfrac{1}{2}mv_f^2

             v_f = \sqrt{2gh_i}

                   = \sqrt{2gL(1- cos\theta)}

                   = \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}

                          = 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the  = 3 m/s

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2

          gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2

             v_f = \sqrt{v_1^2+2gh_i}

             v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}

                       v_f= 11.29 m/s

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