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A block of wood of mass \(55.0 \mathrm{~g}\) floats in a swimming pool, oscillating up and down in simple harmonic motion with a frequency of \(3.00 \mathrm{~Hz}\). a) What is the value of the effective spring constant of the water? b) A partially filled water bottle of almost the same size and shape as the block of wood but with mass \(250 . g\) is placed on the water's surface. At what frequency will the bottle bob up and down?

Short Answer

Expert verified
Answer: The water bottle will bob up and down at a frequency of 1.40 Hz.

Step by step solution

01

Find the angular frequency for the block of wood

The angular frequency (ω) of an oscillating object can be represented by: ω = 2πf where f is the frequency. For the block of wood, we were given f = 3.00 Hz. Plugging this value into the equation, we can find ω: ω = 2π(3.00 Hz) = 18.85 rad/s
02

Calculate the effective spring constant of the water

To find the effective spring constant, we can use the equation for an oscillating object: ω² = k/m where ω is the angular frequency, k is the spring constant, and m is the mass. We have the angular frequency and mass of the block of wood, so we can find the spring constant k: k = m(ω²) = (0.055 kg)(18.85 rad/s)² = 19.37 N/m The effective spring constant of the water is 19.37 N/m.
03

Calculate the angular frequency for the water bottle

The spring constant remains the same for the water bottle, as it is a property of the water. We can use the same equation as in Step 2 to find the angular frequency for the water bottle: ω² = k/m_water_bottle ω = sqrt(k/m_water_bottle) = sqrt((19.37 N/m)/(0.25 kg)) = 8.83 rad/s
04

Calculate the frequency of the oscillating water bottle

Since we have found the angular frequency for the water bottle, we can calculate its oscillation frequency by rearranging the equation from Step 1: f_water_bottle = ω / (2π) = (8.83 rad/s) / (2π) = 1.40 Hz The water bottle will bob up and down at a frequency of 1.40 Hz.

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Key Concepts

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

Angular Frequency
The angular frequency in simple harmonic motion is a fundamental concept that links the time and spatial aspects of oscillation. Imagine the back and forth movement of a pendulum, or the bobbing of an object on water, just like our wooden block. The angular frequency, denoted by the Greek letter \( \omega \), describes how fast this oscillation occurs in terms of radians per second. It is calculated using the formula \( \omega = 2\pi f \) where \( f \) stands for the frequency of oscillation expressed in hertz (Hz).

In everyday terms, angular frequency helps us understand how many full cycles, including all 360 degrees of rotation (which equals 2\pi radians), an oscillating object completes in a second. For the block of wood with a frequency of 3.00 Hz, its angular frequency turns out to be \( 18.85 \text{ rad/s} \), meaning it completes nearly 19 radians of its path every second.

Recognizing the significance of \(\omega\), we can analyze other properties of oscillating systems, as it is also intricately linked to the system's inherent characteristics like the effective spring constant.
Effective Spring Constant
The effective spring constant, commonly represented by \( k \), is akin to a measure of the 'stiffness' or 'springiness' of a system in simple harmonic motion. It's like assessing the rigidity or flexibility of a material without actually bending it. For an object floating and oscillating on water, like our wooden block and water bottle, the effective spring constant connects the force exerted by the water to the displacement of the object.

We can derive this constant from the equation \( \omega^{2} = k/m \) where \( m \) is the mass of the oscillating object. In the case of the block, with its given mass and angular frequency, we calculated the effective spring constant of the water to be \( 19.37 \text{ N/m} \). This tells us that for each meter the block is displaced from its equilibrium position, the water exerts a restorative force of approximately 19.37 newtons to bring it back to equilibrium.

Understanding the effective spring constant is essential when considering different masses on the same medium, as it remains consistent. Its value is intrinsic to the medium—in this case, the water—rather than the object experiencing the harmonic motion.
Oscillation Frequency
The oscillation frequency, which is expressed in hertz (Hz), signifies how many oscillation cycles an object completes in a second. It reflects the rhythmic nature of objects in simple harmonic motion, like a child on a swing going to and fro. Frequency provides a straightforward description of an object's motion that is easy to observe and measure.

From our previous discussions, we understand that the angular frequency is related to the oscillation frequency by the equation \( f = \omega / (2\pi) \). For the partially filled water bottle, we used its mass and the effective spring constant of the water to find an angular frequency of \( 8.83 \text{ rad/s} \) and, through conversion, determined its oscillation frequency to be approximately 1.40 Hz.

Knowing the oscillation frequency helps in predicting the motion of objects subjected to periodic forces and is vital in numerous applications, from designing clocks to analyzing waves. It's a measure directly observable in the real world, making it especially valuable for establishing a clear and practical understanding of the principles governing harmonic motion.

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Most popular questions from this chapter

A small cylinder of mass m can slide without friction on a shaft that is attached to a turntable, as shown in the figure. The shaft also passes through the coils of a spring with spring constant \(k\), which is attached to the turntable at one end, and to the cylinder at the other end. The equilibrium length of the spring (unstretched and uncompressed) matches the radius of the turntable; thus, when the turntable is not

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A mass \(m=1.00 \mathrm{~kg}\) in a spring-mass system with \(k=\) \(1.00 \mathrm{~N} / \mathrm{m}\) is observed to be moving to the right, past its equi. librium position with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\) at time \(t=0 .\) a) Ignoring all damping, determine the equation of motion. b) Suppose the initial conditions are such that at time \(t=0\), the mass is at \(x=0.500 \mathrm{~m}\) and moving to the right with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine the new equation of motion. Assume the same spring constant and mass.

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