What is the phase difference between two simple harmonic motions represented by $x_{1} = A sin (ω t + \frac{π}{6})$ and $x_{2} = A cos (ω t)$ ?

\frac{π}{6}

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\frac{π}{3}

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\frac{π}{2}

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\frac{2 π}{3}

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Solution

The correct option is B $\frac{π}{3}$
Given,
$x_{1} = A sin (ω t + \frac{π}{6})$
$x_{2} = A cos (ω t)$
$x_{2} = A sin (ω t + \frac{π}{2})$

Phase difference:
$△ ϕ = ϕ_{2} - ϕ_{1}$

$△ ϕ = \frac{π}{2} - \frac{π}{6}$

$△ ϕ = \frac{3 π - π}{6} = \frac{π}{3}$

Suggest Corrections

Similar questions

x_{1} = a cos (ω t + (\frac{π}{6}))

x_{2} = a cos (ω t + \frac{π}{3})

If $x = a s i n [ω t + \frac{π}{6}] a n d x^{'} = a c o s ω t$ , then what is the phase difference between the two waves?

y_{1} = a sin [(\frac{π}{2}) t + ϕ]

y_{2} = b sin [(\frac{2 π}{3}) t + ϕ]

x_{1} = A sin (ω t + \frac{π}{6})

x_{2} = A cos ω t

y_{1} = 0.1 s i n (100 π t + \frac{π}{3})

y_{2} = 0.1 c o s π t

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