Home
Class 12
PHYSICS
A solid sphere of radius R is floating i...

A solid sphere of radius `R` is floating in a liquid of density `sigma` with half of its volume submerged. If the sphere is slightly pushed and released , it starts axecuting simple harmonic motion. Find the frequency of these oscillations.

Text Solution

AI Generated Solution

The correct Answer is:
To find the frequency of oscillations of a solid sphere floating in a liquid, we can follow these steps: ### Step 1: Understand the equilibrium condition When the sphere is floating, the buoyant force acting on it is equal to its weight. The buoyant force \( F_b \) can be expressed as: \[ F_b = \sigma \cdot V_s \cdot g \] where \( \sigma \) is the density of the liquid, \( V_s \) is the volume of the submerged part of the sphere, and \( g \) is the acceleration due to gravity. ### Step 2: Determine the volume submerged Given that half of the sphere is submerged, the volume submerged \( V_s \) is: \[ V_s = \frac{1}{2} \cdot V \] where \( V \) is the total volume of the sphere. The volume of a sphere is given by: \[ V = \frac{4}{3} \pi R^3 \] Thus, the submerged volume becomes: \[ V_s = \frac{1}{2} \cdot \frac{4}{3} \pi R^3 = \frac{2}{3} \pi R^3 \] ### Step 3: Calculate the weight of the sphere The weight \( W \) of the sphere can be expressed as: \[ W = \rho_s \cdot V \cdot g \] where \( \rho_s \) is the density of the sphere. Substituting the volume of the sphere: \[ W = \rho_s \cdot \frac{4}{3} \pi R^3 \cdot g \] ### Step 4: Set up the equilibrium condition At equilibrium, the buoyant force equals the weight of the sphere: \[ \sigma \cdot \frac{2}{3} \pi R^3 \cdot g = \rho_s \cdot \frac{4}{3} \pi R^3 \cdot g \] Cancelling common terms, we find: \[ \sigma \cdot \frac{2}{3} = \rho_s \cdot \frac{4}{3} \] From this, we can derive the density of the sphere: \[ \rho_s = \frac{\sigma}{2} \] ### Step 5: Analyze the displacement and the restoring force When the sphere is displaced downward by a small distance \( x \), the increase in submerged volume \( \Delta V \) is: \[ \Delta V = A \cdot x \] where \( A \) is the cross-sectional area of the sphere at the water surface, which is given by: \[ A = \pi R^2 \] Thus, the increase in submerged volume becomes: \[ \Delta V = \pi R^2 \cdot x \] ### Step 6: Calculate the increase in buoyant force The increase in buoyant force \( \Delta F_b \) due to the additional submerged volume is: \[ \Delta F_b = \sigma \cdot \Delta V \cdot g = \sigma \cdot \pi R^2 \cdot x \cdot g \] ### Step 7: Relate the restoring force to the mass of the sphere The restoring force acting on the sphere when displaced is: \[ F = -\Delta F_b = -\sigma \cdot \pi R^2 \cdot g \cdot x \] This is analogous to Hooke's law, where \( F = -k x \) and \( k \) is the effective spring constant: \[ k = \sigma \cdot \pi R^2 \cdot g \] ### Step 8: Find the mass of the sphere The mass \( m \) of the sphere is: \[ m = \rho_s \cdot V = \frac{\sigma}{2} \cdot \frac{4}{3} \pi R^3 = \frac{2}{3} \pi R^3 \sigma \] ### Step 9: Calculate the frequency of oscillation The frequency \( f \) of the oscillation is given by: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \] Substituting the values of \( k \) and \( m \): \[ f = \frac{1}{2\pi} \sqrt{\frac{\sigma \cdot \pi R^2 \cdot g}{\frac{2}{3} \pi R^3 \sigma}} = \frac{1}{2\pi} \sqrt{\frac{3g}{2R}} \] ### Final Result Thus, the frequency of the oscillations is: \[ f = \frac{1}{2\pi} \sqrt{\frac{3g}{2R}} \]
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    RESONANCE ENGLISH|Exercise Exercise- 2, PART - II|1 Videos
  • SIMPLE HARMONIC MOTION

    RESONANCE ENGLISH|Exercise Exercise- 2, PART - III|12 Videos
  • SIMPLE HARMONIC MOTION

    RESONANCE ENGLISH|Exercise Exercise- 1, PART - II|36 Videos
  • SEMICONDUCTORS

    RESONANCE ENGLISH|Exercise Exercise 3|88 Videos
  • TEST PAPERS

    RESONANCE ENGLISH|Exercise PHYSICS|784 Videos

Similar Questions

Explore conceptually related problems

A wooden cube (density of wood 'd' ) of side 'l' flotes in a liquid of density 'rho' with its upper and lower surfaces horizonta. If the cube is pushed slightly down and released, it performs simple harmonic motion of period 'T' . Then, 'T' is equal to :-

A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic montion angular frequency omega . If the radius of the bottle is 2.5 cm then omega is close to: (density of water = 10^(3) kg//m^(3) )

A cubical wood piece of mass 13.5 gm and volume 0.0027 m^3 float in water. It is depressed and released, then it executes simple harmonic motion. Calculate the period of oscillation ?

A solid cube of side a and density rho_(0) floats on the surface of a liquid of density rho . If the cube is slightly pushed downward, then it oscillates simple harmonically with a period of

A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed and then released, find the time period of the resulting simple harmonic motion of the object.

Figure shown a container having liquid of variable density. The density of liquid veriesas rho=rho_(0)(4-(3h)/(h_(0))) . Here, h_(0) and rho_(0) are constants and h is measured from bottom of the container. A solid block of small dimensions whose density is (5)/(2) rho_(0) and mass m is released from bottom of the tank. Prove that the block will execute simple harmonic motion. Find the frequency of oscillation. .

A solid cube floats in a liquid of density rho with half of its volume submerged as shown. The side length of the cube is a, which is very small compared to the size of the container. Now a liquid of density (rho)/(3) (immiscible with the liquid already in the container) is poured slowly into the container. The column of the liquid of density (rho)/(3) that must be poured so that the cube is fully submerged (with its top surface coincident with the surface of the poured liquid) is :

A cylindrical block of wood of mass M is floating n water with its axis vertica. It is depressed a little and then released. Show that the motion of the block is simple harmonic and find its frequency.

A solid sphere of radius r is floating at the interface of two immiscible liquids of densities rho_(1) and rho_(2)(rho_(2) gt rho_(1)) , half of its volume lying in each. The height of the upper liquid column from the interface of the two liquids is h. The force exerted on the sphere by the upper liquid is (atmospheric pressure = p_(0) and acceleration due to gravity is g):

A U-tube contains a liquid to a height of 9.8 cm in one of the limbs. The tube is vertical. If the liquid on one side is depressed a little and released its motion in the tube is simple harmonic motion. Find its period.

RESONANCE ENGLISH-SIMPLE HARMONIC MOTION -Exercise- 2, PART - I
  1. The bob of a simple pendulum of length L is released at time t = 0 fro...

    Text Solution

    |

  2. The period of small oscillations of a simple pendulum of length l if i...

    Text Solution

    |

  3. A simple pendulum , a physical pendulum, a torsional pendulum and a sp...

    Text Solution

    |

  4. A rod of mass M and length L is hinged at its one end and carries a pa...

    Text Solution

    |

  5. A particle moves on the X-axis according to the equation x=x0 sin^2ome...

    Text Solution

    |

  6. The amplitide of a particle due to superposition of following S.H.Ms. ...

    Text Solution

    |

  7. Two particles P and Q describe S.H.M. of same amplitude a, same freque...

    Text Solution

    |

  8. A street car moves rectilinearly from station A to the next station B ...

    Text Solution

    |

  9. A particle is oscillating in a stright line about a centre of force O,...

    Text Solution

    |

  10. Assuming all the surfaces to be smoth, if the time period of motion of...

    Text Solution

    |

  11. A particle of mass m is attached with three springs A,B and C of equal...

    Text Solution

    |

  12. In the figure shown mass 2m is at rest and in equilibrium. A particle ...

    Text Solution

    |

  13. For given spring mass system, if the time period of small oscillations...

    Text Solution

    |

  14. For the arrangement shown in figure, the spring is initially compresse...

    Text Solution

    |

  15. A 1kg block is executing simple harmonic motion of amplitude 0.1 m on ...

    Text Solution

    |

  16. The period of oscillation of a simple pendulum of length L suspended f...

    Text Solution

    |

  17. Figure shown the kinetic energy K of a pendulum versus. its angle thet...

    Text Solution

    |

  18. The bob of a simple pendulum executes SHM in water with a period t. Th...

    Text Solution

    |

  19. A solid sphere of radius R is floating in a liquid of density sigma wi...

    Text Solution

    |

  20. If the angular frequency of small oscillations of a thin uniform verti...

    Text Solution

    |