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SHM of a Floating Object. An object with height \(h\)mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and the cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section 14.7).

Short Answer

Expert verified
The position of the bottom of the object from the liquid surface at equilibrium is \( \frac{M}{\rho A} \), the additional displacement due to the force \( F \) is \( \frac{F}{\rho g A} \), and the period of the ensuing SHM is \( 2\pi \sqrt{\frac{M}{\rho g A}} \).

Step by step solution

01

Equilibrium Position

From the principle of floatation, at equilibrium, the weight of the liquid displaced equals the weight of the floating object. Applying this principle, the equation is given by \( \rho g A L = Mg \), where \( L \)is the distance from the surface of the liquid to the bottom of the floating object and \( g \) is the acceleration due to gravity. Solving this for \( L \), we get \( L = \frac{Mg}{\rho g A} = \frac{M}{\rho A} \).
02

Position Under External Force

With the downward force \( F \), the equation is \( \rho g A (L + d) = Mg + F \), where \( d \) is extra displacement downwards. Solving this for \( d \), we get \( d = \frac{F}{\rho g A} \).
03

Calculation of SHM Period

The restoring force in this case of oscillation is due to buoyancy, which varies linearly with displacement. Thus, the motion is simple harmonic. The period \( T \) of this SHM is given by \( T = 2\pi \sqrt{\frac{m}{k}} \), where \( m \) is the mass of the body and \( k = \rho g A \) is the effective spring constant in this case. So, the period becomes \( T = 2\pi \sqrt{\frac{M}{\rho g A}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyancy and Equilibrium
Understanding the concept of buoyancy and equilibrium is essential in the study of fluids and how objects behave when submerged or floating. Buoyancy is the upward force exerted by a fluid that opposes the weight of an object immersed in it. Equilibrium occurs when the buoyant force is equal to the object's weight, causing it to float without sinking or rising.

Consider an object floating in a liquid. At equilibrium, the weight of the liquid displaced by the submerged part of the object must equal the weight of the object. This relationship is derived from Archimedes' principle and is critical for understanding floating bodies. The equation representing this balance is given by \( \rho g A L = Mg \) where \( L \) is the vertical distance to the bottom of the floating object from the liquid surface, \( A \) is the cross-sectional area of the object, and \( g \) is the acceleration due to gravity. By rearranging this equation, we can solve for \( L \) to find the equilibrium position.
Period of Oscillation
The period of oscillation is a fundamental concept in studying simple harmonic motion (SHM). It represents the time required for an oscillating object to complete one full cycle of its motion. For a floating object that begins to oscillate after being disturbed from its equilibrium position, the period of oscillation is tied to the physical characteristics of the object and the fluid.

The period \( T \) of an object in SHM can be determined using the formula \( T = 2\pi \sqrt{\frac{m}{k}} \) where \( m \) is the oscillating object's mass and \( k \) is the spring constant, which, in the context of buoyancy, relates to the change in buoyant force with displacement. In the scenario of a floating object oscillating in a liquid, \( k \) is given by \( \rho g A \). Hence, we can use this relationship to calculate the period of the oscillation as \( T = 2\pi \sqrt{\frac{M}{\rho g A}} \) which ties the natural frequency of the oscillation to the mass of the object, the density of the fluid, and the cross-sectional area of the object.
Buoyant Force in SHM
In situations involving SHM, such as a floating object that is slightly displaced from its equilibrium position, the buoyant force plays a role analogous to the force exerted by a spring. When the object is pushed down into the liquid, the increased volume of displaced fluid leads to a larger buoyant force, which acts to restore the object to its original position.

The quantity \( k \) in the SHM formula represents the rate at which the buoyant force changes with displacement of the object, akin to the spring constant in Hooke's Law. For a floating object in SHM, \( k \) is determined by the fluid's density \( \rho \) and the cross-sectional area \( A \) of the object, as well as the acceleration due to gravity \( g \) since \( k = \rho g A \). This interrelationship ensures that buoyancy can provide the restoring force necessary for SHM to occur.
Principle of Flotation
The principle of floatation states that a floating object displaces an amount of fluid equal to its own weight. When an object is floating freely, it settles at a position where the gravitational force pulling it downward is balanced by the buoyant force pushing it upward. The volume of the displaced fluid will change if an additional force is applied to the object, such as pushing it further into the fluid or allowing it to rise.

This principle can be applied to determine how much of an object will remain submerged when placed in a fluid. When a force is applied to a floating object, leading it to reach a new equilibrium, we can calculate the additional submerged volume by the change in the depth \( d \) using the formula \( d = \frac{F}{\rho g A} \) where \( F \) is the force applied. Upon the sudden removal of this force, the object will tend to oscillate around its original equilibrium position, governed by the principles of buoyancy and SHM.

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Most popular questions from this chapter

A \(40.0 \mathrm{~N}\) force stretches a vertical spring \(0.250 \mathrm{~m}\). (a) What mass must be suspended from the spring so that the system will oscillate with a period of \(1.00 \mathrm{~s} ?\) (b) If the amplitude of the motion is \(0.050 \mathrm{~m}\) and the period is that specified in part (a), where is the object and in what direction is it moving \(0.35 \mathrm{~s}\) after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is \(0.030 \mathrm{~m}\) below the equilibrium position, moving upward?

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced \(0.120 \mathrm{~m}\) from its equilibrium position and released with zero initial speed, then after \(0.800 \mathrm{~s}\) its displacement is found to be \(0.120 \mathrm{~m}\) on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

A \(0.500 \mathrm{~kg}\) mass on a spring has velocity as a function of time given by \(v_{x}(t)=-(3.60 \mathrm{~cm} / \mathrm{s}) \sin [(4.71 \mathrm{rad} / \mathrm{s}) t-(\pi / 2)] .\) What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is \(0.090 \mathrm{~m},\) it takes the block \(2.70 \mathrm{~s}\) to travel from \(x=0.090 \mathrm{~m}\) to \(x=-0.090 \mathrm{~m} .\) If the amplitude is doubled, to \(0.180 \mathrm{~m},\) how long does it take the block to travel (a) from \(x=0.180 \mathrm{~m}\) to \(x=-0.180 \mathrm{~m}\) and (b) from \(x=0.090 \mathrm{~m}\) to \(x=-0.090 \mathrm{~m} ?\)

BIO Weighing Astronauts. This procedure has been used to "weigh" astronauts in space: A \(42.5 \mathrm{~kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes \(1.30 \mathrm{~s}\) to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes \(2.54 \mathrm{~s}\) for one cycle. What is the mass of the astronaut?

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