Chapter 9, Problem 1RQ

Which of the following is not a right way to express the sinusoid A cos ωt?

1. (a) A cos 2π ft
2. (b) A cos(2πt/T)
3. (c) A cos ω(t − T)
4. (d) A sin(ωt − 90°)

To determine

Choose the correct option which is not a right way to express the sinusoid $A\cos \omega t$.

Answer to Problem 1RQ

The correct option is $\underset{\_}{\left(\text{d}\right)A\sin \left(\omega t-90°\right)}$.

Explanation of Solution

Given data:

The sinusoid function is $A\cos \omega t$.

Calculation:

Option (a):

Given that,

$v\left(t\right)=A\cos \omega t$        (1)

Write the expression to calculate the angular frequency.

$\omega =2\pi f$

Here,

$f$ is the value of the frequency.

Substitute $2\pi f$ for $\omega$ in equation (1) to find $v\left(t\right)$.

$v\left(t\right)=A\cos \left(2\pi f\right)t$        (2)

Therefore, the given sinusoid is also expressed as $A\cos 2\pi ft$.

Option (b):

Write the expression to calculate the frequency.

$f=\frac{1}{T}$

Here,

$T$ is the number of periods.

Substitute $\frac{1}{T}$ for $f$ in equation (2) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=A\cos 2\pi \left(\frac{1}{T}\right)t\\ =A\cos \left(\frac{2\pi t}{T}\right)\end{array}$

Therefore, the given sinusoid is also expressed as $A\cos \left(\frac{2\pi t}{T}\right)$.

Option (c):

The sinusoid function $A\cos \omega t$ can also be expressed as $A\cos \omega \left(t-T\right)$ when the periodic function undergoes a time shifting property.

Option (d):

The equation (1) is,

$\begin{array}{c}v\left(t\right)=A\cos \omega t\\ =A\sin \left(\omega t+90°\right)\left\{\because \cos \omega t=\sin \left(\omega t+90°\right)\right\}\end{array}$

Therefore, the given sinusoid is not expressed as $A\sin \left(\omega t-90°\right)$.

From above, it is clear that the option (d) is correct and the options (a), (b) and (c) are incorrect.

Conclusion:

Thus, the correct option is $\underset{\_}{\left(\text{d}\right)A\sin \left(\omega t-90°\right)}$.