# Frequency problem (static equilibrium)

• toesockshoe
In summary, the lower end of a uniform beam is attached to a vertical wall with a frictionless pivot. The beam makes a 62° angle with the wall and is held in place by a horizontal wire. The wire's length and mass are 4.98-m and 0.737-kg, respectively, while the beam's weight is 349-N. The speed of sound is 344 m/s and when the wind blows, the wire vibrates in its 4th overtone. In attempting to find the tension force and speed of vibration, calculations were made using equations such as T=Ialpha and flambda=v, but errors were encountered.
toesockshoe

## Homework Statement

The lower end of a uniform beam is attached to a vertical wall by a frictionless pivot. The beam extends away from the wall and upward, making a 62° angle with the wall, and it is held in place by a horizontal wire attached from its upper end to the wall. The wire's length and mass are 4.98-m, 0.737-kg and the beam's weight is 349-N. The speed of sound is 344 m/s. When the wind blows, the wire vibrates in its 4th overtone

T=Ialpha
flambda = v

## The Attempt at a Solution

First I wanted to find the tension force so I can find v...

$v = \sqrt {\frac{F_{t}}{\mu}}$
Because the system is in static equilibrium I set $\tau_{F_{g}} = \tau_{T}$ ... $\frac{m_{b}gL}{2} * sin(\theta) = TLcos(\theta)$ ... $T = \frac{m_{b}g}{2}tan(\theta)$

Thus, $v= \sqrt{\frac{\frac{m_{b}g}{2}tan(\theta)}{\frac{0.737}{4.98}}} = 9.456$
$\lambda_{5} = \frac{2L}{5}$ ... solving for $f* \frac{2L}{5} = 9.456$, I get the frequency is 4.747... which isn't the correct answer... can someone tell me where I'm going wrong? Thanks.

toesockshoe said:

## Homework Statement

The lower end of a uniform beam is attached to a vertical wall by a frictionless pivot. The beam extends away from the wall and upward, making a 62° angle with the wall, and it is held in place by a horizontal wire attached from its upper end to the wall. The wire's length and mass are 4.98-m, 0.737-kg and the beam's weight is 349-N. The speed of sound is 344 m/s. When the wind blows, the wire vibrates in its 4th overtone

T=Ialpha
flambda = v

## The Attempt at a Solution

First I wanted to find the tension force so I can find v...

$v = \sqrt {\frac{F_{t}}{\mu}}$
Because the system is in static equilibrium I set $\tau_{F_{g}} = \tau_{T}$ ... $\frac{m_{b}gL}{2} * sin(\theta) = TLcos(\theta)$ ... $T = \frac{m_{b}g}{2}tan(\theta)$

Thus, $v= \sqrt{\frac{\frac{m_{b}g}{2}tan(\theta)}{\frac{0.737}{4.98}}} = 9.456$
Check the value of v.

ehild said:
Check the value of v.

bloody hell.

## 1. What is the frequency problem in static equilibrium?

The frequency problem in static equilibrium refers to the issue of determining the natural frequency of a system at rest. This is typically a challenge because the system is not in motion, making it difficult to measure or calculate the frequency.

## 2. How is the frequency of a system in static equilibrium determined?

The frequency of a system in static equilibrium can be determined using various methods, such as mathematical modeling, experimental testing, or theoretical analysis. These approaches involve considering the system's physical characteristics, such as mass, stiffness, and damping, to calculate or measure the natural frequency.

## 3. Why is it important to determine the frequency of a system in static equilibrium?

Determining the frequency of a system in static equilibrium is crucial for understanding its stability and behavior. The natural frequency of a system determines how it will respond to external forces and disturbances. It also helps in designing and optimizing the system for specific applications.

## 4. What are some common challenges in solving the frequency problem in static equilibrium?

One of the main challenges in solving the frequency problem in static equilibrium is the lack of accurate and complete data on the system's properties. Another challenge is the complexity of the system, which may require advanced mathematical techniques to determine the natural frequency accurately.

## 5. How can the frequency problem in static equilibrium be overcome?

The frequency problem in static equilibrium can be overcome by using advanced analytical and computational techniques, such as finite element analysis, to model and simulate the system's behavior. Experimental testing and validation can also help in determining the natural frequency of the system accurately.

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