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lisov135 [29]
3 years ago
10

If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat

or change?
Physics
1 answer:
Maksim231197 [3]3 years ago
6 0

Answer:

If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2

Explanation:

We know that in a simple harmonic oscillator the maximum speed is given by

    v_{max} = Aw

  Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to w\\ of the oscillation .

  Since  w = 2 \pi f

      v_{max}^{|}/v_{max} = 2f/f = 2

  Where v_{max}^{|} is the maximum speed when frequency is doubled .

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A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=
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Answer:

<em>a) 0.67 rad/sec in the clockwise direction.</em>

<em>b) 98.8% of the kinetic energy is lost.</em>

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angular speed ω = +5.0 rad/s

For second cylinder

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for second cylinder = Iω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = <em>2 kg-m^2 rad/s</em>

When they are coupled coupled together, their total rotational inertia I_{t} = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = I_{t}w_{f}

where I_{t} is their total rotational inertia

w_{f} = their final angular speed together

Final angular momentum = 3 x w_{f} = 3w_{f}

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 =  3w_{f}

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From the first statement, <em>the direction is clockwise</em>

b) Rotational kinetic energy = \frac{1}{2} Iw^{2}

where I is the rotational inertia

w is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = \frac{1}{2} Iw^{2} = \frac{1}{2}*2*5^{2} = 25 J

for second cylinder = \frac{1}{2}*1*8^{2} = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = \frac{1}{2}I_{t}w^{2} _{f}

where

where I_{t} is the total rotational inertia of the cylinders

w_{f} is final total angular speed of the coupled cylinders

Final kinetic energy =  \frac{1}{2}*3*0.67^{2} = 0.67 J

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percentage = 56.33/57 x 100% = <em>98.8%</em>

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