Question 61: Oscillations

Question

A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y \pi 	imes 10^{-2} \mathrm{~s}$, where the value of $y$ is
(Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$, density of water $=10^3 \mathrm{~kg} / \mathrm{m}^3$ )

Accepted Answer

We can determine the time period of the oscillations by considering that when the cube is depressed by a small displacement, the additional buoyant force provided by the displaced water acts as a restoring force. HereÃ¢ÂÂs a stepÃ¢ÂÂbyÃ¢ÂÂstep explanation:
Define the given values:
Side length of the cube, $$L = 10	ext{ cm} = 0.1	ext{ m}$$.
Mass of the cube, $$m = 10	ext{ g} = 0.01	ext{ kg}$$.
Density of water, $$ho = 10^3\ 	ext{kg/m}^3$$.
Acceleration due to gravity, $$g = 10\ 	ext{m/s}^2$$.
Cross-sectional area of the cube (face area),
$$A = L^2 = (0.1)^2 = 0.01	ext{ m}^2.$$
When the cube is depressed by a small distance $$x$$, the additional volume of water displaced is
$$\Delta V = A\,x.$$
Thus, the additional buoyant force is:
$$F_b = ho\,g\,\Delta V = ho\,g\,A\,x.$$
This force acts in the upward (restoring) direction. Notice that the force is proportional to the displacement $$x$$, which is the hallmark of simple harmonic motion (SHM).
The effective spring constant, $$k$$, associated with this SHM is:
$$k = ho\,g\,A.$$
The period of oscillation for SHM is given by:
$$T = 2\pi \sqrt{\frac{m}{k}},$$
which, upon substituting for $$k$$, becomes:
$$T = 2\pi \sqrt{\frac{m}{ho\,g\,A}}.$$
Substitute the given values into the expression:
$$T = 2\pi \sqrt{\frac{0.01}{1000 	imes 10 	imes 0.01}}.$$
Calculate the denominator:
$$1000 	imes 10 	imes 0.01 = 100,$$
so,
$$T = 2\pi \sqrt{\frac{0.01}{100}} = 2\pi \sqrt{0.0001}.$$
Since,
$$\sqrt{0.0001} = 0.01,$$
we have:
$$T = 2\pi 	imes 0.01 = 0.02\pi\ 	ext{s}.$$
The problem states that the time period is given by:
$$y \pi 	imes 10^{-2}\ 	ext{s}.$$
Writing our result in the same form:
$$0.02\pi\ 	ext{s} = 2\pi 	imes 10^{-2}\ 	ext{s}.$$
We see that $$y = 2.$$
Therefore, the correct answer is:
Option A: 2.

Answer

2

Answer

4

Answer

1

Answer

6

Oscillations

Choose the correct answer: