The phase difference between two SHMy1-10sin10- t-3 and y2 -12sin8- t-4 at t-0-5s is-

y_1=10sin(10π t+π/3) Phase 1 =(10π t+π/3) When t=0.5 Phase 1 =(5π+π/3)......(i) y_2=12 sin(8π+π/4) when t=0.5 Phase 2 =(4π t+π/4) .....(ii) phase differ ...

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Standard XIII

Physics

Phase and the General Equation

The phase dif...

Question

The phase difference between two SHM

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

$\frac{11 π}{12}$

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$\frac{13 π}{12}$

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$π$

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$\frac{17 π}{12}$

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Solution

The correct option is B
$\frac{13 π}{12}$

$y_{1} = 10 s i n (10 π t + \frac{π}{3})$

Phase 1 $= (10 π t + \frac{π}{3})$

When t=0.5

Phase 1 $= (5 π + \frac{π}{3})$ ......(i)

$y_{2} = 12 s i n (8 π + \frac{π}{4})$

when t=0.5

Phase 2 $= (4 π t + \frac{π}{4})$ .....(ii)

phase difference=phase 1-phase 2= $5 π + \frac{π}{3} - 4 π - \frac{π}{4}$

$= π + \frac{π}{12}$

$= \frac{13 π}{12}$

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The phase difference between two SHM y1=10sin(10π t+π3) and y2=12sin(8π t+π4) at t=0.5s is:

The correct option is B 13π12 y1=10sin(10π t+π3) Phase 1 =(10π t+π3) When t=0.5 Phase 1 =(5π+π3)......(i) y2=12 sin(8π+π4) when t=0.5 Phase 2 =(4π t+π4) .....(ii) phase difference=phase 1-phase 2=5π+π3−4π−π4 =π+π12 =13π12

The phase difference between two SHM

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