Two SHMs are represented by $y = a sin (w t - k x)$ and $y = b cos (w t - k x)$ . The phase difference between the two is

\frac{π}{2}

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\frac{π}{4}

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\frac{π}{6}

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\frac{3 π}{4}

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Solution

The correct option is A $\frac{π}{2}$
$c o s Θ = s i n (90 - Θ)$
$y = a s i n (ω t - k x)$
$y = b c o s (ω t - k x)$ $= b c o s (\frac{π}{2} - (ω t - k x))$
It is clearly seen that the there is a phase difference of $\frac{π}{2}$ between the two SHMs

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y_{1} = A_{1} sin (w t - k x)

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