Question

Mass–spring systems are used as tuned mass dampers to diminish the vibrations of the balconies of a performing arts center. The oscillation frequency of the TMD is 6.85 Hz, the oscillating mass is 142 kg, and the oscillation amplitude is 4.86 cm.

A) What is the spring constant?
B) What is the maximum speed of the mass?
C) What is the maximum accelerations of the mass?

Answer 1

A)
`k=2.63*10^(5) N/m`.

B)
`v=2.10m/s`

C)
`a=90.0m/s^(2)`

Step-by-step explanation:

This problem is a simple harmonic motion problem. The equation of motion for the SHM is:

`(d^(2)x)/(dt^(2))+\omega^(2)x=0`,

where x is the displacement of the mass about its point of equilibrium, t is time, and
`\omega` is the angular frequency.

A)

First, we need to remember that

`\omega^(2)=(k)/(m)`,

where k is the spring constant, and m is the mass.

From here we can simply solve for k, so

`k=\omega^(2)m`.

Now, we need to make use of an equation that relates the frequency and angular frequency. The equetion is

`\omega=2\pi \\u`,

where
`\\u` is the frequency. This leads us to

`k=(2\pi \\u)^(2)m`,

`k=142(2*6.85*\pi)^(2)`,

`k=2.63*10^(5) N/m`,

B) In simple harmonic motion, the velocity behaves as follow:

`v=\omega Acos(\omega t)` (this is obtained by solving the equation of motion of the mass for the displacement x and take the derivative),

where A is the amplitude of the motion. Since we want the maximum value for the speed, we make
`cos(\omega t)=1` (this because cosine function goes from -1 to 1). With this, the maximum speed is simply

`v = \omega A\\v=(2\pi \\u)A\\v=(2*6.85*\pi)*0.0486\\v=2.10m/s`

C) Here we are going to use the equation of motion of SHM

`(d^(2)x)/(dt^(2))+\omega^(2)x=0`,

we know that

`a=(d^(2)x)/(dt^(2))` , where a is the acceleration,

`a+\omega^(2)x=0\\a=-\omega^(2)x`

in this case, x goes from -A to A, so for a to be maximum we need that
`x=-A` ,and we get

`a=-\omega^(2)(-A)\\a=\omega^(2)A\\a=(2\pi \\u)^(2)A\\a=(2*6.85*\pi)^(2)(0.0486)\\a=90.0m/s^(2)`