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

Answer: The spring  of the spring is 25 N/m.

Explanation:

Mass of the body = 25 g= 0.025 kg (1 kg = 1000 g)

Oscillation is 4 sec = 20

Oscillation in 1 sec =$\frac{20}{4}=5$

Frequency of the vibration of the spring = $5 s^{-1}=5 Hz$

Force constant can be calculated bu using the relation between the frequency and, mass and spring constant 'k'

$Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}$

$5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}$

$k=24.649 N/m\approx 25 N/m$

The spring  of the spring is 25 N/m.