An oscillating block of mass 250 g takes 0.2 sec to move between the endpoints of the motion, which are 50 cm apart. Find the frequency and amplitude of the motion. what is the force constant of the…

Solution

Given Data:

• Mass of the block, $m=250\text{ g}=0.25\text{ kg}$
• Time taken to move between endpoints, $t=0.15\text{ s}$
• Distance between endpoints, $d=40\text{ cm}=0.40\text{ m}$

(a) Frequency of the motion

In Simple Harmonic Motion (SHM), the endpoints of the motion correspond to the maximum positive and maximum negative displacements (extreme positions). The time taken to move from one extreme position to the other is equal to half of the time period ($T/2$).

$$\frac{T}{2}=0.15\text{ s}$$ $$T=2\times 0.15\text{ s}=0.30\text{ s}$$

The frequency ($f$) is the reciprocal of the time period:

$$f=\frac{1}{T}$$ $$f=\frac{1}{0.30}$$ $$f=3.33\text{ Hz}$$

Frequency $f=3.33\text{ Hz}$

(b) Amplitude of the motion

The amplitude ($A$) is defined as the maximum displacement from the equilibrium position. The distance between the two endpoints is equal to twice the amplitude ($2A$).

$$2A=40\text{ cm}$$ $$A=\frac{40}{2}\text{ cm}$$ $$A=20\text{ cm}$$

Converting to SI units: $$A=0.20\text{ m}$$

Amplitude $A=0.20\text{ m}$ (or 20 cm)

(c) Force constant of the spring

The force constant ($k$) of the spring is related to the mass ($m$) and the angular frequency ($\omega$) by the formula:

$$\omega =\sqrt{\frac{k}{m}}⟹k=m{\omega }^2$$

First, calculate the angular frequency $\omega$:

$$\omega =2\pi f$$ $$\omega =2\pi (3.33)$$ $$\omega \approx 20.93\text{ rad/s}$$

Alternatively, using $\omega =\frac{2\pi }{T}$: $$\omega =\frac{2\pi }{0.30}=\frac{20.944}{3}\approx 20.94\text{ rad/s}$$

Now, substitute the values into the equation for $k$:

$$k=m{\omega }^2$$ $$k=0.25\times (20.94{)}^2$$ $$k=0.25\times 438.48$$ $$k=109.62\text{ N/m}$$

Force constant $k\approx 109.6\text{ N/m}$