Question

calculate the amplitude, angular frequency, time period, frequency & intial phase for the simple harmonic oscillation given below
$a.y = 0.3sin(40πt)$
$b.y = 2 \cos(πt)$
$c.y = 3sin(2πt - 1.5)$
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Answer 1

Answer:

The position of a particle at any time 't' in simple harmonic motion is given by the formula

$x(t)=A\sin \left(\omega t+\phi \right)$

Explanation:

• The position of a particle at any time 't' in simple harmonic motion is given by the formula

        $x(t)=A\sin \left(\omega t+\phi \right)$

        where

       $A$- the amplitude of the oscillation

       ω- angular frequency

       Φ- the initial phase

• The time period of oscillation is

        $T=\frac{2\pi }{w}$

• The frequency of oscillation

      $f=1/T$

for oscillations given in the question,

1)   $y=0.3\sin \left(40t\right)$

Compared with the above equation,

the amplitude of the oscillation $A=0.3$

the angular frequency ω$=40$

the initial phase Ф$=0$

Time period $T=2\pi /40=0.157s$

the frequency $f=1/0.157=6.36Hz$

2) $y=2\cos \left(t\right)$

the amplitude $A=2$

the angular frequency ω$=1$

the initial phase Ф$=0$

the time period $T=2\pi /1=6.28s$

the frequency $f=0.159Hz$

3)$y=3sin(2t-1.5)$

the amplitude $A=3$

the angular frequency ω$=2$

the initial phase Ф$=1.5$

time period $T=2\pi /2=3.14s$

frequency $f=1/3.14=0.31Hz$