Chapter 2: Q1P (page 77)

Show that if the line through the origin and the point z is rotated $90 °$ about the origin, it becomes the line through the origin and the point iz. This fact is sometimes expressed by saying that multiplying a complex number byrotates it through $90 °$ . Use this idea in the following problem. Let $z = a e^{i ω t}$ be the displacement of a particle from the origin at time t. Show that the particle travels in a circle of radius a at velocity $v = a ω$ and with acceleration of magnitude directed toward the centre $v^{2} / a$ of the circle.

Short Answer

Expert verified

It has been proved.

$v = a ω A = a ω^{2}$

Step by step solution

Given Information.

The given expression is, $z = a e^{i ω t}$ .

Meaning of rectangular form.

Represent the complex number in rectangular form means writing the given complex number in the form of x + iy in which x is the real part and y is the imaginary part.

Change the angle and find equation.

Consider $z = e^{(i θ)}$

Change the angle by adding $\frac{π}{2}$ .

$θ_{2} = θ + \frac{π}{2}$

Find the new number.

$z = r e^{(i θ_{2})} z = r e^{[(θ + π / 2) i]} z = r e^{(θ i)} . e^{(π i / 2)} z = r e^{(θ i)} [\cos (π / 2) + i \sin (π / 2)] z = r i e^{(θ i)} z = z i$

Find the velocity.

Differentiate the equation with respect to time to find the velocity.

$v = \frac{d z}{d t} v = \frac{d}{d t} [a e^{(i ω t)}] v = a i ω e^{(i ω t)} v = ω i [a e^{(i ω t)}] v = ω i z$

Find the magnitude.

$|v| = |ω i z| |v| = |ω i| . |z| |v| = ω a$

Find the acceleration.

Differentiate the velocity with respect to time to find the acceleration.

$v = \frac{d z}{d t} v = \frac{d}{d t} [ω i a e x p (i ω t)] v = - a ω^{2} e x p (i ω t) v = ω^{2} [a e x p (i ω t)] v = ω^{2} i$

Find the magnitude.

$|A| = |ω^{2} i z| |A| = |ω i| . |z| |A| = ω^{2} a$

Therefore, it has been proved.

$v = a ω A = a ω^{2}$

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Short Answer

Step by step solution

Given Information.

Meaning of rectangular form.

Change the angle and find equation.

Find the velocity.

Find the acceleration.

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