The phase difference between two SHM y1=10 sin (10 π t+(π/3)) and y2=12 sin (8 π t+(π/4)) at t=0.5 s is

Question

The phase difference between two SHM y1=10 sin (10 π t+(π/3)) and y2=12 sin (8 π t+(π/4)) at t=0.5 s is (A) (11 π/12) (B) (13 π/12) (C) π (D) (17 π/12)

Tardigrade Admin · Accepted Answer

Phase for 1st SHM: φ1=10 π t+(π/3) Phase for 2nd SHM: φ2=8 π t+(π/4) Phase difference: Δ φ=φ1-φ2=2 π t+(π/12) Put t=0.5 s Δ φ=2 π(0.5)+(π/12)=(13 π/12)

Q. The phase difference between two SHM $y_{1} = 10 s in$
$(10 π t + \frac{π}{3})$ and $y_{2} = 12 sin (8 π t + \frac{π}{4})$ at $t = 0.5 s$ is

$\frac{11 π}{12}$

$\frac{13 π}{12}$

$π$

$\frac{17 π}{12}$

Phase for 1st SHM: $ϕ_{1} = 10 π t + \frac{π}{3}$
Phase for 2nd SHM: $ϕ_{2} = 8 π t + \frac{π}{4}$
Phase difference: $Δ ϕ = ϕ_{1} - ϕ_{2} = 2 π t + \frac{π}{12}$
Put $t = 0.5 s$
$Δ ϕ = 2 π (0.5) + \frac{π}{12} = \frac{13 π}{12}$

Q. The phase difference between two SHM y1​=10sin (10πt+3π​) and y2​=12sin(8πt+4π​) at t=0.5s is

1211π​

1213π​

π

1217π​

Phase for 1st SHM: ϕ1​=10πt+3π​ Phase for 2nd SHM: ϕ2​=8πt+4π​ Phase difference: Δϕ=ϕ1​−ϕ2​=2πt+12π​ Put t=0.5s Δϕ=2π(0.5)+12π​=1213π​

Q. The phase difference between two SHM $y_{1} = 10 s in$
$(10 π t + \frac{π}{3})$ and $y_{2} = 12 sin (8 π t + \frac{π}{4})$ at $t = 0.5 s$ is

$\frac{11 π}{12}$

$\frac{13 π}{12}$

$π$

$\frac{17 π}{12}$

Phase for 1st SHM: $ϕ_{1} = 10 π t + \frac{π}{3}$
Phase for 2nd SHM: $ϕ_{2} = 8 π t + \frac{π}{4}$
Phase difference: $Δ ϕ = ϕ_{1} - ϕ_{2} = 2 π t + \frac{π}{12}$
Put $t = 0.5 s$
$Δ ϕ = 2 π (0.5) + \frac{π}{12} = \frac{13 π}{12}$