Question

A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , where x is in cm and t is in seconds. What is the frequency of the oscillation?

Answer 1

Answer:

1/2 Hz

Explanation:

A simple harmonic motion has an equation in the form of

$x(t) = Acos(\omega t - \phi)$

where A is the amplitude, $\omega = 2\pi f$ is the angular frequency and $\phi$ is the initial phase.

Since our body has an equation of  x = 5cos(π t + π/3) we can equate $\omega = \pi$ and solve for frequency f

$2\pi f = \pi$

f = 1/2 Hz

Answer 2

Answer:

0.5Hz

Explanation:

The general equation of the displacement, x, of a body undergoing simple harmonic motion at a given point in time (t) is given by;

x = A cos (ωt ± ∅)  --------------------------(i)

where;

A = amplitude of the wave

ω = angular velocity of the wave

∅ = phase constant of the wave

From the question;

x = 5cos(π t + π/3)      -----------------------------(ii)

Comparing equations (i) and (ii), the following deductions among others can be made;

A = 5cm

ω = π

But the angular velocity (ω) of the wave is related to its frequency (f) as follows;

ω = 2 π f        --------------------(iii)

Substitute the value of ω = π  into equation (iii) as follows;

π = 2 π f

Divide through by π;

1 = 2f

Solve for f;

f = 1/2

f = 0.5

Frequency (f) is measured in Hz. Therefore, the frequency of the oscillation is 0.5Hz