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Jlenok [28]
4 years ago
13

Question 5. Our results support the idea that if left to freely oscillate, a system will vibrate at a natural frequency that dep

ends on the system itself, not on the initial push or stimulus that we impart. Based on this reasoning, consider a beam made out of iron and an otherwise identical beam made out of aluminum. How should the natural frequency of vibration of these two beams compare

Physics
1 answer:
Gekata [30.6K]4 years ago
3 0

Answer:

(b) In ideal condition we neglect mass of spring but in real springs mass of spring adds another factor to its time period.

since we are adding a factor of mass to the system, and frequency being inversely proportional to squared root of mass, we can come to a general conclusion that it effectively reduces the natural frequency .

Explanation:

kindly check the attachment for explanation.

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Complete answer

A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100rad/s2. After making 2844 revolutions, its angular speed is 140rad/s

(a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?

Answer:

a) 126.59 radians per second

b) 134.1 seconds

Explanation:

We can use the rotational kinematic equations for constant angular acceleration.

a) For a) let’s use:

\omega^{2}=\omega_{0}^{2}+2\alpha\varDelta\theta (1)

with \omega_{0} the initial angular velocity, \omega the final angular velocity, \alpha the angular acceleration and \Delta \thetathe revolutions on radians (2844 revolutions = 17869.38 radians). Solving (1) for initial velocity:

\sqrt{\omega^{2}-2\alpha\varDelta\theta}=\omega_{0}

\omega_{0}^2=\sqrt{(140)^2 -(2)(0.100)(17869.38)=126.59 \frac{rad}{s}}

b) Knowing those values, we can use now the kinematic equation

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t=\frac{\omega-\omega_0}{\alpha}=\frac{140-126.59}{0.1}

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