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A mass m at the end of a spring vibrates with a frequency
of0.88 Hz. When an additional 680-g mass is added to m, the
frequencyis 0.60 Hz. What is the value of m?

Respuesta :

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

[tex]F = -kx[/tex]

And Newton's Second Law also states that

[tex]F = ma = m\frac{d^2x}{dt^2}[/tex]

Combining two equations yields

[tex]a = -\frac{k}{m}x[/tex]

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

[tex]\omega = \sqrt{\frac{k}{m}}[/tex]

And given that ω = 2πf

the relation between frequency and mass becomes

[tex]f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex].

Let's apply this to the variables in the question.

[tex]0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg[/tex]

Lanuel

The value of the initial mass (m) is equal to 0.5906 kg.

Given the following data:

  • Initial frequency = 0.88 Hz.
  • Additional mass = 680 grams to kg = 0.68 kg
  • New frequency = 0.60 Hz

To determine the value of the initial mass, m, we would apply the following formula:

[tex]F = \frac{1}{2\pi} \sqrt{\frac{k}{m} }[/tex]

Where:

  • F is the frequency.
  • m is the mass.
  • k is the spring constant.

For mass (m):

[tex]0.88 = \frac{1}{2\pi} \sqrt{\frac{k}{m} }[/tex]    ...equation 1.

For additional mass:

[tex]0.60 = \frac{1}{2\pi} \sqrt{\frac{k}{m + 0.68} }[/tex]   ...equation 2.

Next, we would divide eqn. 1 by eqn. 2:

[tex]\frac{0.88}{0.60} = \frac{ \frac{1}{2\pi} \sqrt{\frac{k}{m}} }{\frac{1}{2\pi} \sqrt{\frac{k}{m + 0.68} }} \\\\\\1.4667 = \frac{\sqrt{\frac{k}{m}} }{\sqrt{\frac{k}{m + 0.68}} }\\\\[/tex]

Taking the square of both sides, we have:

[tex]2.1512= \frac{\frac{k}{m}}{\frac{k}{m + 0.68}} \\\\2.1512=\frac{k}{m} \times \frac{m + 0.68}{k} \\\\2.1512= \frac{m + 0.68}{m} \\\\2.1512m = m+0.68\\\\2.1512m - m=0.68[/tex]

Simplifying further, we have:

[tex]1.1512m =0.68\\\\m=\frac{0.68}{1.1512}[/tex]

Mass, m = 0.5906 kg

Read more: https://brainly.com/question/14708169

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