Two sinusoidal voltages of the same frequency | vedclass

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(B) From the graph, the time period $T$ (time taken to complete one full cycle) for wave $M$ is $0.4 \, s$.Therefore, the frequency $
u = \frac{1}{T}

Vedclass · Accepted Answer

$2.5 \, Hz, -\pi/2 \, 	ext{radians}$

Vedclass · Answer

(B) From the graph, the time period $T$ (time taken to complete one full cycle) for wave $M$ is $0.4 \, s$.Therefore, the frequency $
u = \frac{1}{T} = \frac{1}{0.4} = 2.5 \, Hz$.Wave $M$ starts at $t = 0$ with zero value and positive slope, representing $\sin(\omega t)$.Wave $N$ starts at $t = 0$ with a negative value, reaching its peak later than $M$. Specifically, $N$ is delayed by a quarter of a cycle compared to $M$.A delay of a quarter cycle corresponds to a phase lag of $\frac{\pi}{2}$ radians.Thus, the phase lead of $N$ over $M$ is $-\frac{\pi}{2}$ radians.

Two sinusoidal voltages of the same frequency are shown in the diagram. What is the frequency, and the phase relationship between the voltages? (Frequency in $Hz$, Phase lead of $N$ over $M$ in radians)

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