Question

An oscillator consists of a block of mass $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{\text{kg}}\mathbf{}$connected to a spring. When it is set into oscillation with amplitude $\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$, the oscillator repeats its motion every $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{\text{s}}$. Find the (a) period, (b) frequency, (c) angular frequency, (d) spring constant, (e) maximum speed, and (f) magnitude of the maximum force on the block from the spring.

Short answer

1. The period of the oscillations is$0.500\text{s}$ .
2. The frequency of oscillations is$2.00\text{Hz}$ .
3. The angular frequency of oscillations is $12.56\text{rad}/\text{s}$.
4. The spring constant is $79.0\text{N}/\text{m}$.
5. The maximum speed of the oscillations is $4.40\text{m}/\text{s}$.
6. The maximum force on the block from the spring is$27.65\text{N}$ .

Step-by-step solution

Stating the given data

1. Mass of the body,$\mathit{\text{m}}=0.500\text{kg}$
2. Amplitude of the oscillator,$A=35\text{cm or}0.35\text{m}$
3. Time at which motion repeats itself,$T=0.5\text{s}$ .

Understanding the concept of motion

Using the formula of the period, we can find the frequency of oscillations. Using the relation between angular velocity and frequency, we can find the angular frequency of the oscillations. Then using the relation between angular frequency and spring constant, we can find the spring constant.

Using the formula maximum of speed, we can find the maximum speed of the oscillations.

Using the formula of spring force, we can find the maximum force on the block from the spring.

Frequency of a body in oscillation,

$f=\frac{1}{T}$ ......(i)

The angular frequency of a body in oscillation,

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

Angular frequency of a spring,

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

The maximum speed of a body in oscillation,

$v_m=\omega x_m$......(iv)

The force applied on a spring,

$F_m=k{x}_m$ ......(v)

a) Calculation of period

Period of the oscillation is the time required to complete one oscillation.

Therefore, the period of the oscillation is $T=0.500\text{s}$, because the motion repeats every $0.500\text{s}$.

b) Calculation of frequency

Using equation (i) and $T=0.500\text{s}$, we get the frequency of oscillation as

$\begin{array}{rcl}f& =& \frac{1}{0.500\text{s}}\\ & =& 2.00\text{Hz}\end{array}$

Therefore, the frequency of oscillations is $2.00\text{Hz}$.

c) Calculation of angular frequency

Using equation (ii) and $\text{f}=2.00\text{Hz}$, we get the angular frequency of oscillation as,

$\begin{array}{rcl}\omega & =& 2\pi \left(2.00\text{Hz}\right)\\ & =& 12.56\text{rad}/\text{s}\end{array}$

Therefore, the angular frequency of oscillations is$12.56\text{rad}/\text{s}$ .

d) Calculation of spring constant

Using equation (iii), we get the spring constant of oscillation as:

$k=m{\omega }^2$

As

$\omega =12.56\text{rad}/\text{s}$and $m=0.500\text{kg}$

$\begin{array}{rcl}k& =& \left(0.500\text{kg}\right){\left(12.6\frac{\text{rad}}{\text{s}}\right)}^2\\ & =& 78.88\text{N}/\text{m}\\ & \approx & \text{79.0 N/m}\\ & & \end{array}$

Therefore, the spring constant is $79.0\text{N}/\text{m}$.

e) Calculation of maximum speed

Using equation (iv) and the given values, we get the velocity of oscillation as:

$\begin{array}{rcl}{v}_m& =& \left(12.56\frac{\text{rad}}{\text{s}}\right)\left(0.350\text{m}\right)\\ & =& 4.396\text{m}/\text{s}\\ & \approx & 4.40\text{m/s}\end{array}$

Therefore, the maximum speed is $4.40\text{m}/\text{s}$.

f) Calculation of maximum force

Using equation (iv) and the given values, we get the velocity of oscillation as:

$\begin{array}{rcl}{F}_m& =& \left(79.0\frac{\text{N}}{\text{m}}\right)\left(0.350\text{m}\right)\\ & =& 27.65\text{N}\end{array}$

Therefore, the maximum force is$27.65\text{N}$