Question

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 1

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:

$\omega = 2\pi \cdot f$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$f$ - Frequency, measured in hertz.

Given that $f = 4\,hz$, the angular frequency of the system is:

$\omega = 2\pi \cdot (4\,hz)$

$\omega \approx 25.133\,\frac{rad}{s}$

Now, the angular frequency can be obtained in terms of spring constant and mass. That is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

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

$m$ - Mass, measured in kilograms.

The spring constant is now cleared:

$k = \omega^{2}\cdot m$

If $\omega = 25.133\,\frac{rad}{s}$ and $m = 0.20\,kg$, the spring constant is:

$k = \left(25.133\,\frac{rad}{s} \right)^{2}\cdot (0.20\,kg)$

$k = 126.334\,\frac{N}{m}$

The spring has a constant of 126.334 newtons per meter.