91Ó°ÊÓ

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,$T=0.5\text{s}$ .

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,

$\mathrm{Ó\neg }=2\mathrm{Ï€}f$ ......(ii)

Angular frequency of a spring,

$\mathrm{Ó\neg }=\sqrt{\frac{k}{m}}$......(iii)

The maximum speed of a body in oscillation,

$v_m=\mathrm{Ó\neg }{x}_m$......(iv)

The force applied on a spring,

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

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 $0.500\text{s}$.

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 $2.00\text{Hz}$.

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

$\begin{array}{rcl}\mathrm{Ó\neg }& =& 2\mathrm{Ï€}\left(2.00\text{Hz}\right)\\ & =& 12.56\text{rad}/\text{s}\end{array}$

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

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

$k=m{\mathrm{Ó\neg }}^2$

As

$\mathrm{Ó\neg }=12.56\text{rad}/\text{s}$and $m=0.500\text{kg}$

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

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

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

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

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$27.65\text{N}$