Question

When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of the spring? include units, no spaces. round to two significant figures?

Answer 1

The system makes 20 complete vibrations in 4.0 s, so its frequency is
$f= \frac{20}{4.0 s}=5.0 Hz$

While the angular frequency is
$\omega = 2 \pi f = 2 \pi (5.0 Hz)=31.4 rad/s$

For a simple harmonic motion, the angular frequency is also equal to
$\omega = \sqrt{ \frac{k}{m} }$
where k is the spring constant, while m=25 g=0.025 kg is the mass attached to the spring. Using the value of the angular frequency we calculated before, we can find the value of k:
$k=\omega^2 m=(31.4 rad/s)^2 (0.025 kg)=25 N/m$

Answer 2

The spring constant of the spring in two significant figures is 25 N/m.

The given parameters:

• mass attached, m = 25 g
• number of revolutions, N = 20
• time of motion, t = 4.0 s

The angular speed of the spring is calculated as follows;

$\omega = \frac{2\pi N}{t} \\\\\omega = \frac{2\pi \times 20}{4} \\\\\omega = 31.42 \ rad/s$

The spring constant is calculated as follows;

$\omega = \sqrt{\frac{k}{m} } \\\\(\omega)^2 = \frac{k}{m} \\\\k =\omega ^2 m\\\\k = (31.42)^2 \times 0.025\\\\k = 24.68 \ N/m\\\\k \approx 25 \ N/m$

Thus, the spring constant of the spring in two significant figures is 25 N/m.

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