Question

Equation of simple harmonic motion Y= a sin (4Πt + Thita) its frequency is
1) 2. 2) 1/2 3) 2Π 4) Π/2

Answer 1

Answer:

Given:

Equation of. SHM has been provided as follows

$y = a \sin( 4\pi t + \theta)$

To find:

Frequency of the particle performing Oscillations.

Concept:

Frequency is the number of Oscillations made in 1 complete second.

Mathematically , we can say that :

$\boxed{ \large{ \red{freq. = \dfrac{ \omega}{2\pi}}}}$

where, ω is the angular Frequency.

Calculation:

Comparing with a standard Equation of SHM ,

$\boxed{ \red{y = a \sin( \omega t + \phi)}}$

ω = 4π, and θ = φ = initial phase.

So frequency is as follows :

$\therefore \: freq. = \dfrac{4\pi}{2\pi}$

$\implies \: freq. = 2 \: hz$

So final answer is 2 Hz frequency.