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Two point masses of 3.0 kg and 1.0 kg are attached to opposite ends of a horizontal spring whose spring constant is 300 Nm?1 as shown in the figure. The natural vibration frequency of the system is about: (a) 4 Hz (b) 3 H (c) 2 Hz (d) 1 Hz

Shane Macguire , 11 Years ago
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Askiitians Tutor Team

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To determine the natural vibration frequency of a system consisting of two point masses attached to a spring, we can use the formula for the frequency of a mass-spring system. The formula for the natural frequency \( f \) in hertz (Hz) is given by:

Frequency Formula

The formula is:

f = \frac{1}{2\pi} \sqrt{\frac{k}{m_{eff}}}

Understanding the Components

In this equation:

  • f is the natural frequency in hertz.
  • k is the spring constant, which in this case is 300 N/m.
  • m_{eff} is the effective mass of the system.

Calculating the Effective Mass

For two masses attached to a spring, the effective mass \( m_{eff} \) can be calculated using the formula:

m_{eff} = \frac{m_1 \cdot m_2}{m_1 + m_2}

Here, \( m_1 = 3.0 \, \text{kg} \) and \( m_2 = 1.0 \, \text{kg} \). Plugging in these values:

m_{eff} = \frac{3.0 \cdot 1.0}{3.0 + 1.0} = \frac{3.0}{4.0} = 0.75 \, \text{kg}

Substituting into the Frequency Formula

Now that we have the effective mass, we can substitute \( k \) and \( m_{eff} \) into the frequency formula:

f = \frac{1}{2\pi} \sqrt{\frac{300}{0.75}}

Calculating the value inside the square root:

\frac{300}{0.75} = 400

Now, taking the square root:

\sqrt{400} = 20

Finally, substituting back into the frequency formula:

f = \frac{1}{2\pi} \cdot 20

Calculating this gives:

f \approx \frac{20}{6.2832} \approx 3.18 \, \text{Hz}

Choosing the Closest Answer

From the options provided, the closest value to our calculated frequency of approximately 3.18 Hz is:

  • (b) 3 Hz

Thus, the natural vibration frequency of the system is about 3 Hz. This calculation illustrates how the mass and spring constant interact to determine the frequency of oscillation in a mass-spring system.

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