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Chapter 10: Problem 50

(a) Sketch a graph of \(x(t)=A \sin \omega t\) (the position of an object in SHM that is at the equilibrium point at \(t=0\) ). (b) By analyzing the slope of the graph of \(x(t),\) sketch a graph of $v_{x}(t) .\( Is \)v_{x}(t)$ a sine or cosine function? (c) By analyzing the slope of the graph of \(v_{x}(t),\) sketch \(a_{x}(t)\) (d) Verify that \(v_{x}(t)\) is \(\frac{1}{4}\) cycle ahead of \(x(t)\) and that \(a_{x}(t)\) is \(\frac{1}{4}\) cycle ahead of \(v_{x}(t) .\) (W) tutorial: sinusoids)

Short Answer

Expert verified
Question: Sketch the graphs for the position, velocity, and acceleration functions for simple harmonic motion. Verify the relationships between their phases. Answer: In simple harmonic motion, we have the position function \(x(t) = A \sin(\omega t)\). The graph of this function is a sine wave with amplitude A and period \(T = \frac{2\pi}{\omega}\). The velocity function is given by \(v_x(t) = A\omega \cos(\omega t)\), and the graph resembles a cosine wave with amplitude A\(\omega\). The acceleration function is \(a_x(t) = -A\omega^2 \sin(\omega t)\) which is equal to \(- \omega^2 x(t)\), having a similar graph as the position function but inverted and scaled by a factor of -\(\omega^2\). The relationships between the phases can be verified as follows: the velocity function \(v_x(t)\) is \(\frac{1}{4}\) cycle ahead of the position function \(x(t)\) because the cosine function reaches its maximum value \(\frac{1}{4}\) cycle ahead of the sine function. The acceleration function \(a_x(t)\) is \(\frac{1}{4}\) cycle ahead of the velocity function \(v_x(t)\) because a \(\sin(\omega t)\) function reaches its minimum (most negative) value at \(\frac{1}{4}\) cycle later than a \(\cos(\omega t)\) function reaching its maximum value.

Step by step solution

01

Part (a): Sketch the graph of \(x(t) = A \sin(\omega t)\).

To draw the graph of \(x(t) = A \sin(\omega t)\), first recognize that it is a sine function with an amplitude of A and angular frequency of \(\omega\). The graph starts at the equilibrium point and oscillates between maximum and minimum values equal to A and -A, respectively. The period of the function is given by \(T = \frac{2\pi}{\omega}\).
02

Part (b): Analyzing the slope of \(x(t)\) to find \(v_x(t)\) and identify its function type.

To find the velocity function \(v_x(t)\), we need to determine the derivative \(\frac{dx(t)}{dt}\) of the position function \(x(t)=A \sin(\omega t)\). We have: \(\frac{dx(t)}{dt} = A\omega \cos(\omega t)\). This is a cosine function with amplitude A\(\omega\) and angular frequency \(\omega\). The sketch of \(v_x(t)\) is similar to a cosine function, starting from the maximum value A\(\omega\) and oscillating between maximum and minimum values of A\(\omega\) and -A\(\omega\), respectively.
03

Part (c): Analyzing the slope of \(v_x(t)\) to find \(a_x(t)\).

To find the acceleration function \(a_x(t)\), we need to determine the derivative \(\frac{dv_x(t)}{dt}\) of the velocity function \(v_x(t)=A\omega \cos(\omega t)\). We have: \(\frac{dv_x(t)}{dt} = -A\omega^2 \sin(\omega t)\) which is equal to \(- \omega^2 x(t)\). The graph of the acceleration function is the same as the position function but inverted and scaled by a factor of -\(\omega^2\).
04

Part (d): Verify relationships between position, velocity, and acceleration functions.

To show that \(v_x(t)\) is \(\frac{1}{4}\) cycle ahead of \(x(t)\), note that the cosine function reaches its maximum value \(\frac{1}{4}\) cycle ahead of the sine function. Also, the maximum slope of the sine function occurs when the cosine function is at its maximum value. To show that \(a_x(t)\) is \(\frac{1}{4}\) cycle ahead of \(v_x(t)\), recall that \(a_x(t) = -A\omega^2 \sin(\omega t) = -\omega^2 x(t)\). A \(\sin(\omega t)\) function reaches its minimum (most negative) value at \(\frac{1}{4}\) cycle later than a \(\cos(\omega t)\) function reaching its maximum value. Thus, the relationships between \(x(t)\), \(v_x(t)\), and \(a_x(t)\) are confirmed.

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