Question

a simple haromic oscillator is represented by the equation Y=0.4sin(440t+0.61x)​

Answer 1

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A simple harmonic oscillator is represented by the equation : Y = 0.4 Sin (440t+0.61)

Y is in metres

t is in seconds

Then Find the values of :

1) Amplitude

2) Angular frequency

3) Frequency of oscillation

4) Time period of oscillation

5) Initial phase.

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As, we know that

$\Large \implies {\boxed{\red{\sf{Y \: = \: a \: Sin (\omega t \: + \: \varnothing)}}}}$

So, Compare both

We get,

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(1) Amplitude (a) = 0.4 m

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(2) Angular frequency (ω) = 440 Hz

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(3) Frequency of oscillation

We have formula :-

$\Large {\boxed{\red{\sf{\mu \: = \: \frac{\omega}{2 \pi}}}}}$

$\Large \leadsto {\sf{\mu \: = \: \frac{440}{2 \: \times \: \frac{22}{7}}}}$

$\Large \leadsto {\sf{\mu \: = \: \frac{\cancel{440}}{\frac{\cancel{44}}{7}}}}$

$\Large \leadsto {\sf{\mu \: = \: 10 \: \times \: 7}}$

$\Large \implies {\boxed{\red{\sf{\mu \: = \: 70 \: Hz}}}}$

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(4) Time Period

As we know that,

$\Large \hookrightarrow {\sf{T \: = \: \frac{1}{\mu}}}$

⇒ T = 1/70

⇒T = 0.0143 s

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(5) Initial Phase (∅) = 0.16 radians