Step-by-step explanation:
Frequency of a mass-spring system is:
f = √(k/m) / (2π)
Given f = 2/π Hz and m = 20 kg:
2/π Hz = √(k / 20 kg) / (2π)
4 Hz = √(k / 20 kg)
16 s⁻² = k / 20 kg
k = 320 kg/s²
k = 320 N/m
If the 20 kg mass is replaced with an 80 kg mass, the frequency is:
f = √(320 kg/s² / 80 kg) / (2π)
f = √(4 s⁻²) / (2π)
f = 2 Hz / (2π)
f = 1/π Hz
The sprint constant is 320 N/m
And, the frequency is [tex]1\div \pi Hz[/tex]
Since A 20-kilogram mass is attached to a spring. And, the frequency of simple harmonic motion is 2/p cycles/s.
Now we know that
[tex]f = \sqrt (k\div m) \div (2\pi)[/tex]
Here
[tex]f = 2\div \pi Hz[/tex]
and m = 20 kg:
So,
[tex]2\div \pi Hz = \sqrt (k \div 20 kg) / (2\pi)\\\\4 Hz = \sqrt(k \div 20 kg)\\\\16 s^{-2} = k \div 20 kg[/tex]
k = 320 kg/s²
k = 320 N/m
Now the frequency is:
[tex]f = \sqrt(320 kg/s^2 \div 80 kg) \div (2\pi)\\\\f = √(4 s^{-2}) \div (2\pi )\\\\f = 2 Hz \div (2\pi)[/tex]
[tex]f = 1\div \pi Hz[/tex]
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