Respuesta :
Answer:
The spring constant k is [tex]k=480[/tex] N/m.
The frequency if the original mass is replaced with an 120 kilogram mass is [tex]\frac{1}{\pi }[/tex] cycles/s.
Step-by-step explanation:
From the information given:
- A 30 kilogram mass is attached to a spring.
- The frequency of simple harmonic motion is 2/π cycles/s.
and we want to find the spring constant k.
We can use this equation
[tex]\frac{d^2x}{dt^2} +\omega^2x=0[/tex], where [tex]\omega^2=\frac{k}{m}[/tex]
that describe simple harmonic motion.
The general solution of this equation is:
[tex]x(t)=c_1cos(\omega t)+c_2sin(\omega t)[/tex]
From this equation, the frequency of motion, [tex]f=\frac{1}{T}=\frac{\omega}{2\pi }[/tex] is the number of cycles completed every one second, where [tex]T[/tex] is the period of motion and [tex]\omega =\sqrt{\frac{k}{m} }[/tex] is called the circular frequency of the system.
Thus,
[tex]f=\frac{1}{T}=\frac{\omega}{2\pi } =\frac{1}{2\pi } \sqrt{\frac{k}{m} }[/tex]
(a) We know the frequency [tex]f=\frac{2}{\pi }[/tex] cycles/s and the mass [tex]m=30[/tex] kg, so we can solve the above equation for k the spring constant.
[tex]f\cdot \:2\pi =\frac{1}{2\pi }\sqrt{\frac{k}{m}}\cdot \:2\pi \\\\f\cdot \:2\pi =\sqrt{\frac{k}{m}}\\\\\left(f\cdot \:2\pi \right)^2=\left(\sqrt{\frac{k}{m}}\right)^2\\\\4\pi ^2f^2=\frac{k}{m}\\\\k=4\pi ^2f^2m[/tex]
[tex]k=4\pi ^2(\frac{2}{\pi } )^2(30) = 480[/tex] N/m
(b) To find the frequency of simple harmonic motion if the original mass is replaced with an 120 kilogram mass you must use the value of the spring constant that we found in (a) and this equation,
[tex]f=\frac{1}{2\pi \:}\:\sqrt{\frac{k}{m}\:}\\\\f=\frac{1}{2\pi \:}\:\sqrt{\frac{480}{120}\:}\\\\f=\frac{1}{\pi }[/tex]
The frequency is [tex]\frac{1}{\pi }[/tex] cycles/s.
A 30 kilogram mass is attached to a spring. If the frequency of simple harmonic motion is 2/π cycles/s, then the spring constant k is 480N/m and If we replace original mass with an 120 kilogram mass the frequency of simple harmonic motion is 1/π cycles/s
What is harmonic motion?
Harmonic motion is a special type of periodic motion in which the restoring force on the moving object is directly proportional to the magnitude of the object's displacement.
Given that mass=30kg
frequency of simple harmonic motion is 2/π cycles/s
We can find spring constant as
[tex]\frac{d^{2} x}{d t^{2}}+\omega^{2} x=0$\\\\ where $\omega^{2}=\frac{k}{m}[/tex]
General solution of the above equation is:
[tex]$x(t)=c_{1} \cos (\omega t)+c_{2} \sin (\omega t)$[/tex]
[tex]f=\frac{1}{T}=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}[/tex]
[tex]$\begin{aligned}&f \cdot 2 \pi=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \cdot 2 \pi \\&f \cdot 2 \pi=\sqrt{\frac{k}{m}} \\&(f \cdot 2 \pi)^{2}=\left(\sqrt{\frac{k}{m}}\right)^{2} \\&4 \pi^{2} f^{2}=\frac{k}{m} \\&k=4 \pi^{2} f^{2} m \\&k=4 \pi^{2}\left(\frac{2}{\pi}\right)^{2}(30)=480 \mathrm{~N} / \mathrm{m}\end{aligned}$[/tex]
If we replace original mass with an 120 kilogram mass
[tex]$\begin{aligned}&f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \\&f=\frac{1}{2 \pi} \sqrt{\frac{480}{120}} \\&f=\frac{1}{\pi}\end{aligned}$[/tex]
A 30 kilogram mass is attached to a spring. If the frequency of simple harmonic motion is 2/π cycles/s, then the spring constant k is 480N/m and If we replace original mass with an 120 kilogram mass the frequency of simple harmonic motion is 1/π cycles/s
To learn more about Harmonic motion visit :https://brainly.com/question/17315536
