Problem 55
Question
The basic "dimensions" of mechanics are length, mass, and time. Trigonometric functions are unitless because they represent the ratio of two lengths-the units cancel. Arguments of trigonometric functions are angles and so unitless. Likewise, exponential functions are unitless and take unitless arguments. An oscillation about equilibrium is given by \(x(t)=\) \(A \sin (\omega t)\) when \(t\) is measured in seconds and \(x\) in meters. What units do the constants \(A\) and \(\omega\) bear? Use the Chain Rule to explain why \(x^{\prime}(t)\) is dimensionally correct.
Step-by-Step Solution
Verified Answer
A is in meters; \(\omega\) is in \(s^{-1}\). Derivative maintains dimensional correctness with velocity units.
1Step 1: Determine Units of Amplitude A
The function for oscillation is given by \(x(t) = A \sin (\omega t)\). Since \(x(t)\) is measured in meters, the amplitude \(A\), which is the maximum displacement from equilibrium, must also be in meters. Therefore, \(A\) has the unit of length, which is meters \([m]\).
2Step 2: Understand Trigonometric Argument
The term \(\omega t\) inside the sine function must be unitless. Since \(t\) is given in seconds (\([s]\)), \(\omega\) must have the units of reciprocal seconds (\([s^{-1}]\)) or radians per second, so that the product \(\omega t\) is unitless.
3Step 3: Use Chain Rule for Dimensionality of Derivative
The derivative \(x^{\prime}(t)\) with respect to \(t\) is \(A \omega \cos(\omega t)\). When you differentiate \(A \sin(\omega t)\), you use the chain rule: the derivative of \(\sin(\omega t)\) is \(\omega \cos(\omega t)\). This keeps the argument unitless while introducing \(\omega\) from the derivative of \(t\). The overall unit for \(x^{\prime}(t)\) is \(A \omega\), or \(([m][s^{-1}])\), which is meters per second. This is consistent with the unit for velocity, confirming dimensional correctness.
Key Concepts
Trigonometric FunctionsOscillationDerivativesChain Rule
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent are fundamental in mathematics and physics, especially in mechanics. They usually work with angles and are essential in modeling periodic phenomena such as oscillations and waves. What's interesting about trigonometric functions is that they take unitless arguments. This is because the arguments are ratios of two lengths. Therefore, the units cancel, resulting in a dimensionless value.
- Their outputs are also unitless, representing pure numbers instead of measurement dimensions.
- This property makes them pivotal in calculations involving periodicity and harmonics.
Oscillation
Oscillation describes the repetitive back-and-forth motion of an object around an equilibrium position. In mechanics, this concept is well-represented by functions such as \(x(t) = A \sin (\omega t)\). Here,
This periodic motion can show both simple and complex patterns depending on the factors influencing it, such as damping or external forces.
- \(A\) is the amplitude, indicating the maximum extent of the oscillation from the equilibrium point.
- \(\omega\) is the angular frequency, showing how fast the oscillation occurs.
This periodic motion can show both simple and complex patterns depending on the factors influencing it, such as damping or external forces.
Derivatives
In calculus, derivatives are key to understanding changes and rates. They represent the rate of change of a function concerning its variable. For the oscillating function \(x(t) = A \sin (\omega t)\),
- the derivative \(x'(t)\) provides the velocity of the motion at any given time.
- This derivative calculation involves applying the chain rule to find \(A \omega \cos(\omega t)\).
Chain Rule
The chain rule is a crucial concept in calculus for differentiating composed functions. In the context of mechanics and oscillations, it helps in finding the derivative of functions like \(x(t) = A \sin (\omega t)\).
- To differentiate \(\sin(\omega t)\), you first handle the inner function, \(\omega t\).
- The chain rule involves multiplying the derivative of the outer function (like \(\cos\)) by the derivative of the inner function (\(\omega\)).
Other exercises in this chapter
Problem 55
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