A light hollow cube of side length 10 cm and mass 10g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is \( y \pi \times 10^{-2} \) s, where the value of \( y \) is:
(Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \), density of water = \( 10^3 \, \text{kg/m}^3 \))
(Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \), density of water = \( 10^3 \, \text{kg/m}^3 \))
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The Correct Option is A
Approach Solution - 1
To find the time period of oscillations of the cube executing simple harmonic motion in water, we need to apply the principles of fluid mechanics and simple harmonic motion.
The cube is floating in water, meaning its buoyant force balances its weight, which can be understood through Archimedes' principle.
For simple harmonic motion, the time period \( T \) is given by the formula:
\(T = 2\pi \sqrt{\frac{m}{k}}\)
For this floating object:
- m is the mass of the cube.
- k is the effective spring constant related to the buoyant force.
The force due to buoyancy acting on the cube when it is displaced a small distance \( x \) is:
\(F_b = - \rho g A x\)
where:
- \(\rho = 1000 \, \text{kg/m}^3\) (density of water)
- \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity)
- \(A = 0.1 \, \text{m} \times 0.1 \, \text{m} = 0.01 \, \text{m}^2\) (cross-sectional area of the cube)
This force can be associated with an effective spring constant \(k\):
\(k = \rho g A\)
Substitute values:
\(k = 1000 \times 10 \times 0.01 = 100 \, \text{N/m}\)
Mass of the cube \(m\) is 10 g, which is 0.01 kg.
Substitute \( m \) and \( k \) into the time period formula:
\(T = 2\pi \sqrt{\frac{0.01}{100}}\)
\(T = 2\pi \sqrt{0.0001} = 2\pi \times 0.01 = 0.02\pi \, \text{s}\)
This matches the expression \( y \pi \times 10^{-2} \, \text{s}\), where \( y = 2 \).
Hence, the value of \( y \) is 2.
Approach Solution -2
To find the time period of oscillations of the cube, we start by analyzing the oscillations of a floating object. The buoyant force acting on the cube when it is displaced from its equilibrium position provides the restoring force necessary for simple harmonic motion.
The side length of the cube is given as \(10 \, \text{cm}\), which is \(0.1 \, \text{m}\). The cube is floating in water, so the volume of water it displaces is equal to its own volume.
Step 1: Finding the Volume
The volume \( V \) of the cube is:
\( V = (0.1 \, \text{m})^3 = 0.001 \, \text{m}^3 \)
The mass of the cube is given as \(10 \, \text{g}\), which is \(0.01 \, \text{kg}\).
The weight of the cube is \( mg = 0.01 \times 10 = 0.1 \, \text{N} \).
In equilibrium, the weight of the cube is balanced by the buoyant force:
\( F_b = \rho V g = 10^3 \times 0.001 \times 10 = 10 \, \text{N} \)
Since the mass of the cube is much smaller than the buoyant force, the restoring force is due to the change in the buoyant force when the cube is displaced.
The displacement changes the volume of water displaced by an additional amount, \( A x \) (where \( A = \text{cross-sectional area}\) and \( x \) is the displacement).
Step 2: Calculating the force
The effective restoring force on the cube when displaced by \( x \) is:
\( F = -\rho A g x \)
For a cube, the cross-sectional area \( A = 0.1 \times 0.1 = 0.01 \, \text{m}^2 \).
The effective mass \( m_{\text{eff}} \) participating in oscillation is essentially the mass of the displaced fluid:
\( m_{\text{eff}} = \rho V = 10^3 \times 0.001 = 1 \, \text{kg} \)
The equation of motion can be given by:
\( m_{\text{eff}} \frac{d^2x}{dt^2} = -\rho A g x \)
Which simplifies to the formula for simple harmonic motion:
\( \frac{d^2x}{dt^2} = -\left(\frac{\rho A g}{m_{\text{eff}}}\right)x \)
The angular frequency \( \omega \) is then:
\( \omega = \sqrt{\frac{\rho A g}{m_{\text{eff}}}} \)
The time period \( T \) of oscillation is:
\( T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m_{\text{eff}}}{\rho A g}} \)
Substituting the known values:
\( T = 2\pi \sqrt{\frac{1}{10^3 \times 0.01 \times 10}} = 2\pi \times 0.1 \)
Thus, the time period is \( T = 2\pi \times 0.1 = 0.2\pi \) seconds, which can be expressed as \( y\pi \times 10^{-2} \) seconds where \( y = 2 \).
Therefore, the value of \( y \) is 2.
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