Question

In Figure, a block weighing 14.0 N, which can slide without friction on

an incline at angle$\mathbf{40}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}$, is connected to the top of the incline by a massless

spring of unstretched length 0.450 mand spring constant 120 N/m.

a) How far from the top of the incline is the block’s equilibrium point?

b) If the block is pulled slightly down the incline and released, what is the period

of the resulting oscillations?

Short answer

a) The equilibrium point for the block from the top of the inclined plane is 0.525 m

away.

b) Time period of oscillations is 0.686 s , when it is slightly pulled down the incline and

released.

Step-by-step solution

The given data

1. Angle of inclination of the wedge, $\sout{0}={40}^{\circ }$.
2. Weight of the block, W = 14.0 N .
3. Length of the spring, L = 0.45 m .

4. Spring constant, k = 120 N/m .

Understanding the concept of oscillatory motion

By drawing the free body diagram for the block, we can equate the forces in the direction parallel to the inclined plane. After solving the equation, we get the displacement of the spring. When we add the length of the unstretched spring, we get the exact position of the equilibrium point for the block.

Using the weight formula, we can find the mass of the block. Hence, from this and the given data, we can find the period for the oscillations.

Formula:

The force on a spring,F = Kx…….(i)

The time period of a body under oscillations,$\mathrm{T}=2\mathrm{\pi r}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$……(ii)

Weight of a body due to gravity,W = mg……(iii)

(a) Calculation of distance of the equilibrium’s point

Free body diagram for the block

From the diagram we can say that,

$$\begin{gathered} \mathrm{mg}\sin \sout{0}=\mathrm{kx}\left(\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right)\right) \\ \mathrm{w}\sin \sout{0}=\mathrm{kx}\left(\mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right)\right) \\ \mathrm{x}=\frac{\mathrm{w}\sin \sout{0}}{\mathrm{k}} \\ =\frac{14\times \sin {40}^{\circ }}{120} \\ =0.075\mathrm{m} \end{gathered}$$

We know that the length of the unstretched spring is 0.45 m

Hence, the distance of equilibrium point from top of inclined plane

$$\begin{gathered} \mathrm{d}=0.45+0.075 \\ =0.525\mathrm{m} \end{gathered}$$

Hence, the value of the distance is

(b) Calculation of period of resulting oscillations

Using equation (i), we get the mass of block-system as:

$$\begin{gathered} 14=\mathrm{m}\times 9.8 \\ \mathrm{m}=\frac{14}{9.8} \\ =1.43\mathrm{kg} \end{gathered}$$

Using equation (ii) and the given values, the period is given as:

$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{1.43}{120}} \\ =2\mathrm{\pi }\times 0.1091 \\ =0.686\mathrm{s} \end{gathered}$$

Hence, the value of period is0.686s.