A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 8.0 cm. If the mass of the object is 0.20 kg, what is the spring constant
A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 8.0 cm. If the mass of the object is 0.20 kg, what is the spring constant
Answer:
The spring has a constant of 126.334 newtons per meter.
Explanation:
Given that mass-spring system experiment a simple harmonic motion, the angular frequency of the system as a function of frequency is:
[tex]\omega = 2\pi \cdot f[/tex]
Where:
[tex]\omega[/tex] – Angular frequency, measured in radians per second.
[tex]f[/tex] – Frequency, measured in hertz.
Given that [tex]f = 4\,hz[/tex], the angular frequency of the system is:
[tex]\omega = 2\pi \cdot (4\,hz)[/tex]
[tex]\omega \approx 25.133\,\frac{rad}{s}[/tex]
Now, the angular frequency can be obtained in terms of spring constant and mass. That is:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Where:
[tex]k[/tex] – Spring constant, measured in newtons per meter.
[tex]m[/tex] – Mass, measured in kilograms.
The spring constant is now cleared:
[tex]k = \omega^{2}\cdot m[/tex]
If [tex]\omega = 25.133\,\frac{rad}{s}[/tex] and [tex]m = 0.20\,kg[/tex], the spring constant is:
[tex]k = \left(25.133\,\frac{rad}{s} \right)^{2}\cdot (0.20\,kg)[/tex]
[tex]k = 126.334\,\frac{N}{m}[/tex]
The spring has a constant of 126.334 newtons per meter.