Natural frequency of a pendulum being lowered at 2m/s

In summary, the conversation discusses a heavy machine being lowered vertically by a winch and stopping suddenly, causing vibration. The problem can be solved by using Young's modulus for steel to find the amplitude and period of the ensuing vibration.
  • #1
Dustinsfl
2,281
5
A heavy machine weighing \(9810\) N is being lowered vertically down by a winch at a uniform velocity of \(2\) m/s. The steel cable supporting the machine has a diameter of \(0.01\) m. The winch is suddenly stopped when the steel cable's length is \(20\) m. Find the period and amplitude of the ensuing vibration of the machine.

From the question, we know that \(m\ddot{x} = 9810\) N, the pendulum will have a length of \(20\) m, and then we have the diameter of cable and the speed at which it was lowered. I am not sure if the last two are superfluous.

How does one tackle this type of problem?
 
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  • #2
dwsmith said:
A heavy machine weighing \(9810\) N is being lowered vertically down by a winch at a uniform velocity of \(2\) m/s. The steel cable supporting the machine has a diameter of \(0.01\) m. The winch is suddenly stopped when the steel cable's length is \(20\) m. Find the period and amplitude of the ensuing vibration of the machine.

From the question, we know that \(m\ddot{x} = 9810\) N, the pendulum will have a length of \(20\) m, and then we have the diameter of cable and the speed at which it was lowered. I am not sure if the last two are superfluous.

How does one tackle this type of problem?

Use Young's modulus for steel.
 
  • #3
I like Serena said:
Use Young's modulus for steel.

It says that the young's modulus for steel is 200.
\[
\omega_n = \sqrt{\frac{200A}{mL}}
\]
where \(A\) is the cross sectional area and \(L\) is the length. So L = 20 and \(A = 0.000025\pi\). The mass is \(m = 9810/g \approx 1000\).
\[
\omega_n = 0.000886227
\]
This doesn't seem right.
 
  • #4
Young's modulus for steel is 200 GPa.
 
  • #5
I like Serena said:
Young's modulus for steel is 200 GPa.

dwsmith said:
A heavy machine weighing \(9810\) N is being lowered vertically down by a winch at a uniform velocity of \(2\) m/s. The steel cable supporting the machine has a diameter of \(0.01\) m. The winch is suddenly stopped when the steel cable's length is \(20\) m. Find the period and amplitude of the ensuing vibration of the machine.

From the question, we know that \(m\ddot{x} = 9810\) N, the pendulum will have a length of \(20\) m, and then we have the diameter of cable and the speed at which it was lowered. I am not sure if the last two are superfluous.

How does one tackle this type of problem?

From 9810 N you can find the mass of the ball by just dividing 9.81, and mas becomes simply 1000 kg. Now you can find Kinetic Energy of the ball. When the winch is suddenly stopped, the cable gets elongated and a potential energy develops which resist the ball. Now Youngs Modulus gives you the force applied on the cable which comes out as

F = (EA/L)*x=kx where E young's modulus (200 GPa), A area of the cable, L length of the cable, and x is the increment in length.
Elastic potential energy = kx*x/2
Equating with KE, you can get x which gives amplitude.
To find the period apply F=kx or T=2*pi*sqrt(m/k)
In my rough calculation period and amplitude comes about 0.19 sec and 0.035 mt. Best of luck.
 

FAQ: Natural frequency of a pendulum being lowered at 2m/s

What is the natural frequency of a pendulum?

The natural frequency of a pendulum is the rate at which it swings back and forth when displaced from its resting position. It is determined by the length of the pendulum and the force of gravity.

How is the natural frequency of a pendulum calculated?

The natural frequency of a pendulum can be calculated using the formula: f = 1/(2π)√(g/L), where f is the frequency in hertz, g is the acceleration due to gravity (9.8 m/s^2), and L is the length of the pendulum in meters.

What factors affect the natural frequency of a pendulum?

The natural frequency of a pendulum is affected by the length of the pendulum, the force of gravity, and the amplitude (distance it swings from its resting position). Other factors such as air resistance and the mass of the pendulum can also have an impact.

How does lowering a pendulum at 2m/s affect its natural frequency?

The speed at which a pendulum is lowered does not affect its natural frequency. As long as the length and force of gravity remain constant, the natural frequency will not change. However, the amplitude (distance it swings) may be affected by the initial velocity of the pendulum.

Can the natural frequency of a pendulum be changed?

Yes, the natural frequency of a pendulum can be changed by altering the length of the pendulum or the force of gravity acting on it. This can be done by adjusting the position of the pendulum's pivot point or changing its mass. Additionally, factors like air resistance and the amplitude can also impact the natural frequency.

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