Question

A mass m at the end of a spring vibrates with a frequency of 0.72 Hz . When an additional 700 g mass is added to m, the frequency is 0.64 Hz . Part A What is the value of m? Express your answer using two significant figures.

Answer 1

The value of m is 2635.294 grams.

Step-by-step explanation:

Let suppose that mass-spring system has a simple harmonic motion, to this respect the formula for frequency is:

`f = (\omega)/(2\pi)`

Where
`\omega` is the angular frequency, measured in radians per second.

For a mass-spring system under simple harmonic motion, the angular frequency is:

`\omega = \sqrt{(k)/(m) }`

Where:

`k` - Spring constant, measured in newtons per meter.

`m` - Mass, measured in kilograms.

The following equation is obtained after replacing angular frequency in frequency formula:

`f = (1)/(2\pi)\cdot \sqrt{(k)/(m) }`

As this shows, frequency is inversely proportional to the square root of mass. Hence, the following relationship is deducted:

`f_(1)\cdot \sqrt{m_(1)} = f_(2) \cdot \sqrt{m_(2)}`

If
`m_(2) = m_(1) + 700\,g`,
`f_(1) = 0.72\,hz` and
`f_(2) = 0.64\,hz`, the resulting expression is simplified and then initial mass is found after clearing it:

`f_(1) \cdot \sqrt{m_(1)} = f_(2) \cdot \sqrt{m_(1)+700\,g}`

`f_(1)^(2) \cdot m_(1) = f_(2)^(2)\cdot (m_(1) + 700\,g)`

`\left((f_(1))/(f_(2)) \right)^(2)\cdot m_(1) = m_(1) + 700\,g`

`\left[\left((f_(1))/(f_(2))\right)^(2) - 1\right]\cdot m_(1) = 700\,g`

`m_(1) = (700\,g)/(\left((f_(1))/(f_(2)) \right)^(2)-1)`

`m_(1) = (700\,g)/(\left((0.72\,hz)/(0.64\,hz) \right)^(2)-1)`

`m_(1) = 2635.294\,g`

The value of m is 2635.294 grams.