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The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12 degrees to the vertical.
a. with what frequency does it vibrate? assume shm.
b. what is the pendulum bob's speed when it passes through the lowest point of the swing?
c. what is the total energy stored in this oscillation, assuming no losses?

Respuesta :

a. The frequency should be 0.614Hz

b The pendulum bob's speed is 0.443 m/s

c. The total energy stored is 0.0304 J

Conservation of energy:

a. The frequency should be

Here we have to first apply the periodic time

T= 2π√(L/g)

Now the frequency should be

=  1/T (Hz)

T = 2 × 3.14 √ (0.66/9.81)

= 6.28 × √0.0673

1.6289 Seconds

Now

Frequency = 1/T

= f = 1/1.6289

= 0.614 Hz

b)  The speed of the bob is

Here the vertical distance via the height is

h= 0.66 cos 12

h = 0.65 m

Now

Vertical fall at the lowest point  should be

= 0.66 - 0.65

= 0.01 m

Now

energy lost (MgΔh) = KE gained (1/2mv²)

mgh = 1/2mv²

 v² = 2gΔh = 2×9.81 × 0.01

= 0.1962

So, v = 0.443 m/s

c) The total energy is

E = 1/2mv²

  = 1/2 × 0.310 × 0.443²

  = 0.0304 J

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