Two SHMs are given by Y 1 -Asin- 2 t- and Y 2 -Bsin 2- t - 3 t- -The phase difference between these two after -1- sec is

The correct option is D π 6 For phase difference ϕ_1 ϕ_2 = 2π t3 + ϕ #160; π t2 ϕ #160; #160; #160; #160; #160; #160; #160 ...

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Standard XII

Physics

Time Period Independent of Amplitude

Two SHMs are ...

Question

Two SHMs are given by $Y_{1} = A s i n (\frac{π}{2} t + ϕ)$ and $Y_{2} = B s i n (\frac{2 π t}{3} t + ϕ)$ .The phase difference between these two after '1' sec is

π

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\frac{π}{2}

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\frac{π}{4}

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\frac{π}{6}

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Solution

The correct option is D $\frac{π}{6}$
For phase difference

$ϕ_{1} - ϕ_{2} = \frac{2 π t}{3} + ϕ - \frac{π t}{2} - ϕ$

= $\frac{π t}{6}$

SO AT t=1 sec

$ϕ_{1} - ϕ_{2} = \frac{π}{6}$

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The phase difference between two SHM

$y_{1} = 10 s i n (10 π t + \frac{π}{3}) a n d y_{2} = 12 s i n (8 π t + \frac{π}{4}) a t t = 0.5 s i s :$

Q. Equations of a stationary and a travelling waves are as follows,

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and

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S H M

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Q. Equations of a stationary wave and a travelling wave are

y_{1} = a s i n k x c o s ω t

and

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Two SHMs are given by Y1=Asin(π2t+ϕ) and Y2=Bsin(2πt3t+ϕ).The phase difference between these two after '1' sec is

The correct option is D π6For phase differenceϕ1−ϕ2=2πt3+ϕ−πt2−ϕ =πt6SO AT t=1 secϕ1−ϕ2=π6

Two SHMs are given by $Y_{1} = A s i n (\frac{π}{2} t + ϕ)$ and $Y_{2} = B s i n (\frac{2 π t}{3} t + ϕ)$ .The phase difference between these two after '1' sec is

The correct option is D $\frac{π}{6}$
For phase difference

$ϕ_{1} - ϕ_{2} = \frac{2 π t}{3} + ϕ - \frac{π t}{2} - ϕ$

= $\frac{π t}{6}$

SO AT t=1 sec

$ϕ_{1} - ϕ_{2} = \frac{π}{6}$