Question

The motion of a particle is given by y = 9(cm) sin(ωt) + 3(cm) cos(ωt) The motion is


Not an SHM


SHM with amplitude 12 cm


SHM with amplitude


SHM with amplitude 3 cm

Answer 1

Answer:

Motion of a particle is given as :

$y = 9 \sin( \omega t) + 3 \cos( \omega t)$

We need to find out if this particle is undergoing SHM or not. And if yes , what's the Amplitude of Oscillation ?

Talking 3 common , we get :

$y = 3 \{3 \sin( \omega t) + \cos( \omega t) \}$

$= > y = 3( \sqrt{ {3}^{2} + {1}^{2} } ) \{ (\frac{3}{ \sqrt{ {3}^{2} + {1}^{2} } } ) \sin( \omega t) + (\frac{1}{ \sqrt{ {3}^{2} + {1}^{2} } }) \cos( \omega t) \}$

$= > y = 3 \sqrt{10} \{ \frac{3}{ \sqrt{10}} \sin( \omega t) + \frac{1}{ \sqrt{10}} \cos( \omega t) \}$

Let 3/√10 be $\cos(\theta)$ ,

then 1/√10 will be $\sin(\theta)$

Continuing with the equation , we get:

$= > y = 3 \sqrt{10} \{ \cos( \theta) \sin( \omega t) + \sin( \theta) \cos( \omega t) \}$

$= > y = 3 \sqrt{10} \sin( \omega t + \theta)$

So it's an SHM with amplitude of 3√10 cm.

Answer 2

Explanation:

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$\bf \huge \: Question \: \:$

• The motion of a particle is given by y = 9(cm) sin(ωt) + 3(cm) cos(ωt) busy The motion is

• Not an SHM
• SHM with amplitude 12 cm

• SHM with amplitude

• SHM with amplitude 3 cm

_________________________

$\bf \huge \: GIVEN \: \:$

• The motion of a particle is given by y = 9(cm) sin(ωt) + 3(cm) cos(ωt)

_________________________

$\bf \huge \: To\: Find \:$

• What Is motion?

_________________________

considering the Equation,

y= A sin(ωt) + B cos(ωt)

Now let ,cos = a/√a^2+b^2

and sin = b/√a^2+b^2

now Putting the above given the Equation,

________________________

we get :

$\bf \:y = 3 \{3 \sin( \omega t) + \cos( \omega t) \}$

$\bf \:= > y = 3( \sqrt{ {3}^{2} + {1}^{2} } ) \{ (\frac{3}{ \sqrt{ {3}^{2} + {1}^{2} } } ) \sin( \omega t) + (\frac{1}{ \sqrt{ {3}^{2} + {1}^{2} } }) \cos( \omega t) \}$

________________________

Now Using the identify,

$\bf \:= > y = 3 \sqrt{10} \{ \frac{3}{ \sqrt{10}} \sin( \omega t) + \frac{1}{ \sqrt{10}} \cos( \omega t) \}$

Let 3/√10 be

$\bf \:\cos(\theta)$ ,

then 1/√10

$\bf \:\sin(\theta)$

_________________________

we get:

$\bf \:= > y = 3 \sqrt{10} \{ \cos( \theta) \sin( \omega t) + \sin( \theta) \cos( \omega t) \}$

$\bf \: = > y = 3 \sqrt{10} \sin( \omega t + \theta)$

Hope it helps you.