Two simple harmonic motions are represented by the equations y 1 = 0.1sin 100 πt + π 3 and y 2 = 0.1cos 100 πt . The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

ν 1 = d dt y 1 = 0.1 × 100 π cos 100 πt + π 3 ν 2 = d dt y 2 = - 0.1 × π sin πt = 0.1 × π cos πt + π 2 ∴ Δ ϕ = π 3 - π 2 = - π 6

Start Free Mock LoginAlready registered? Login here

Physics - SHM Question with Solution | TestHub

PhysicsSHMMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

Two simple harmonic motions are represented by the equations $y_{1} = 0.1sin (100 πt + \frac{π}{3})$ and $y_{2} = 0.1cos (100 πt)$ . The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is

Options:

Answer:

Solution:

$ν_{1} = \frac{d}{dt} (y_{1}) = (0.1 \times 100 π) cos (100 πt + \frac{π}{3})$

$ν_{2} = \frac{d}{dt} (y_{2}) = (- 0.1 \times π) sin πt$

$= (0.1 \times π) cos (πt + \frac{π}{2})$

$∴$ $Δ ϕ = \frac{π}{3} - \frac{π}{2} = - \frac{π}{6}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...