Two simple harmonic motions are represented by equations y1=4 sin (10 t+φ) ⇒ y2=5 cos 10 t What is the phase difference between their velocities?

Question

Two simple harmonic motions are represented by equations y1=4 sin (10 t+φ) ⇒ y2=5 cos 10 t What is the phase difference between their velocities? (A) φ (B) -φ (C) (φ+(π/2)) (D) (φ-(π/2))

Tardigrade Admin · Accepted Answer

y1 = 4 sin(10 t + φ) y2 = 5 cos 10 t v1 = (dy1/dt) = 40 cos(10 t + φ) v2 = (dy2/dt) = -50 sin 10t = 50 cos(10t + π/2) Phase difference between v1 and v2 = cos(10t + π/2)

Q. Two simple harmonic motions are represented by equations $y_{1} = 4 sin (10 t + ϕ) \Rightarrow y_{2} = 5 cos 10 t$ What is the phase difference between their velocities?

$ϕ$

$- ϕ$

$(ϕ + \frac{π}{2})$

$(ϕ - \frac{π}{2})$

$y_{1} = 4 sin (10 t + ϕ)$
$y_{2} = 5 cos 10 t$
$v_{1} = \frac{d y _{1}}{d t} = 40 cos (10 t + ϕ)$
$v_{2} = \frac{d y _{2}}{d t} = - 50 sin 10 t$
$= 50 cos (10 t + π /2)$
Phase difference between $v_{1}$ and $v_{2} = cos (10 t + π /2)$

Q. Two simple harmonic motions are represented by equations y1​=4sin(10t+ϕ)⇒y2​=5cos10t What is the phase difference between their velocities?

ϕ

−ϕ

(ϕ+2π​)

(ϕ−2π​)

y1​=4sin(10t+ϕ) y2​=5cos10t v1​=dtdy1​​=40cos(10t+ϕ) v2​=dtdy2​​=−50sin10t =50cos(10t+π/2) Phase difference between v1​ and v2​=cos(10t+π/2)

Q. Two simple harmonic motions are represented by equations $y_{1} = 4 sin (10 t + ϕ) \Rightarrow y_{2} = 5 cos 10 t$ What is the phase difference between their velocities?

$ϕ$

$- ϕ$

$(ϕ + \frac{π}{2})$

$(ϕ - \frac{π}{2})$

$y_{1} = 4 sin (10 t + ϕ)$
$y_{2} = 5 cos 10 t$
$v_{1} = \frac{d y _{1}}{d t} = 40 cos (10 t + ϕ)$
$v_{2} = \frac{d y _{2}}{d t} = - 50 sin 10 t$
$= 50 cos (10 t + π /2)$
Phase difference between $v_{1}$ and $v_{2} = cos (10 t + π /2)$