Equation of two S-H-M- x1 - 5 sin 2 - t - - -4- x2 - 5- 2 sin 2- t-cos2- t ratio of amplitude phase difference will be

The correct option is B 1 : 2, 0 We have, #160; X_1=5sin(2nt+n/4) A_1=5 m and ϕ_1=n/4 X_2=5√(2)(sin 2nt+cos 2nt) X_2=5√(2)×√(2)( ...

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Standard XII

Physics

Phasors

Equation of t...

Question

Equation of two S.H.M. $x_{1} = 5 s i n (2 π t + π / 4),$ $x_{2} = 5 \sqrt{2} (s i n 2 π t + c o s 2 π t)$ ratio of amplitude & phase difference will be

2 : 1, 0

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1 : 2, 0

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1 : 2, π / 2

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2 : 1, π / 2

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Solution

The correct option is B $1 : 2, 0$
We have,
$X_{1} = 5 sin (2 n t + n / 4)$
$A_{1} = 5 m$ and $ϕ_{1} = n / 4$
$X_{2} = 5 \sqrt{2} (sin 2 n t + cos 2 n t)$
$X_{2} = 5 \sqrt{2} \times \sqrt{2} (\frac{sin 2 n t}{\sqrt{2}} + \frac{cos 2 n t}{\sqrt{2}})$
$X_{2} = 10 sin (2 n t + π / 4)$
$A_{2} = 10 m$ and $ϕ_{2} = π / 4$
$\frac{A_{1}}{A_{2}} = \frac{5 m}{10 m} = \frac{1}{2}$ and $Δ ϕ = ϕ_{1} - ϕ_{2} = 0$
Option $B$ is correct.

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Equation of two S.H.M. x1=5sin(2πt+π/4), x2=5√2(sin2πt+cos2πt) ratio of amplitude & phase difference will be

The correct option is B 1:2,0We have, X1=5sin(2nt+n/4)A1=5 m and ϕ1=n/4X2=5√2(sin2nt+cos2nt)X2=5√2×√2(sin2nt√2+cos2nt√2)X2=10sin(2nt+π/4)A2=10 m and ϕ2=π/4A1A2=5m10m=12 and Δϕ=ϕ1−ϕ2=0Option B is correct.

Equation of two S.H.M. $x_{1} = 5 s i n (2 π t + π / 4),$ $x_{2} = 5 \sqrt{2} (s i n 2 π t + c o s 2 π t)$ ratio of amplitude & phase difference will be