Question

Question: An oscillator consists of a block attached to a spring (k = 400 N/m). At some time t, the position (measured from the system’s equilibrium location), velocity, and acceleration of the block are, x =0.100 m,v = 13.6 m and a = 123 m/s². Calculate (a) the frequency of oscillation,(b) the mass of the block, and (c) the amplitude of the motion.

Short answer

Answer:

1. The frequency of oscillation is 5.58 Hz .
2. The mass of the block is 0.325 kg.
3. The amplitude of the motion is 0.400 m.

Step-by-step solution

The given data

1. Spring constant of the spring, k= 400 N/m
2. Position at time t, x =0.100 m
3. Velocity at time t, v = - 13.6 m
4. Acceleration at time t, a = 123 m/s²

Understanding the concept of oscillatory motion

Using the formula of acceleration of SHM and using the relation between angular velocity and frequency, we can find the frequency of oscillation.

Using the formula of angular velocity in terms of spring constant and mass, we can find the mass of the block. Then, using the law of energy conservation, we can find the amplitude of the motion.

Formula:

The acceleration of a body in oscillation,

$a=-{\omega }^{2}x$ (i)

The angular frequency of a body in oscillation,

$\omega =2\pi f$ (ii)

The angular frequency of a spring,

$\omega =\sqrt{k/m}$ (iii)

The potential energy of a body in oscillation,

$U=\frac{1}{2}k{x}_m^2$ (iv)

The kinetic energy of a body in motion,

$K=\frac{1}{2}m{v}^2$ (v)

(a) Calculation of frequency of oscillation

Using equation (i) and the given values, we get the angular frequency as:

$$\begin{gathered} \omega =\sqrt{-\frac{a}{x}}(\because a=-123{\text{m/s}}^{\text{2}}\&x=0.100\text{m}) \\ =\sqrt{-\frac{\left(-123\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)}{0.100\text{m}}} \\ =35.07\text{rad/s} \end{gathered}$$

Again using the equation (ii), we get the frequency as:

$$\begin{gathered} f=\frac{\omega }{2\pi } \\ =\frac{35.07\frac{\text{rad}}{\text{s}}}{2\pi } \\ =5.58\text{Hz} \end{gathered}$$

Therefore, thefrequency of oscillation is 5.58 Hz

(b) Calculation for the mass of the oscillator

Using equation (iii), we get the mass as:

$$\begin{gathered} m=\frac{k}{{\omega }^2}(\because k=400\text{N/m}\&\omega =35.07\text{rad/s}) \\ =\frac{400\text{N/m}}{{\left(35.07\frac{\text{rad}}{\text{s}}\right)}^2} \\ =0.325\text{kg} \end{gathered}$$

Therefore, the mass of the block is 0.325 Kg

(c) Calculation of amplitude of the motion

Using equation (iv) & (v), the total energy of the system can be given as:

$\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$

Therefore, according to energy conservation law,

$$\begin{gathered} \frac{1}{2}k{x}_m^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2 \\ {x}_m^2=x^2+\frac{m}{k}{v}^2 \\ {x}_m=\sqrt{x^2+\frac{m}{k}{v}^2} \\ =\sqrt{{\left(0.100m\right)}^2+\frac{\left(0.325\text{kg}\right)}{\left(400\text{N/m}\right)}{\left(-13.6\frac{\text{m}}{\text{s}}\right)}^2} \\ =0.400\text{m} \end{gathered}$$

Therefore, the amplitude of the motion is 0.400 m.