Question

21. A simple harmonic motion is represented by x = 12 sin (10t +0.6) find out the Amplitude,
frequency.
time period and angular frequency if displacement measured in meter and time in seconds.​

Answer 1

Answer:

amplitude=12

frequency=  $10/2pi$

time period =reciprocal of frequency=$2pi/10$

Explanation:

Answer 2

Answer:

Amplitude = 12 m

Frequency = 1.59 Hz

Time period = 0.62 sec

Angular frequency = 10 rad/sec

Explanation:

A simple harmonic motion is represented by,

• x = 12 sin (10t + 0.6)

Equation of SHM ( simple harmonic motion ) is

• x = A sin ( ωt + ϕ )

Here,

A = Amplitude

ω = Angular frequency

ϕ = Initial phase

On comparing both the equations, we get:

Amplitude, A = 12 m

Angular frequency, ω = 10 rad/sec

Initial phase, ϕ = 0.6 rad

As we know,

Angular frequency, ω = 2 π f.         Here f = frequency

so,

f = ω/2π

$f = \frac{10}{2 * 3.14}$ = 1.59 Hz

As we know that frequency is inversely proportional to time period,

so,

$T = \frac{1}{f}$

$T = \frac{1}{1.59}$ = 0.62 sec

RESULT: Amplitude = 12 m

               Frequency = 1.59 Hz

               Time period = 0.62 sec

               Angular frequency = 10 rad/sec