Question

The displacement of a particle is given by X equal to 3 sin 5πt the + 4 cos 5πt then the amplitude of particle is

Answer 1

Answer:

Displacement of a particle is given by

$x = 3 \sin(5\pi t) + 4 \cos(5\pi t)$

In order to find maximum displacement, we need to express this equation in terms of single sine or cosine function.

$= > x = \bigg( \sqrt{ {3}^{2} + {4}^{2} } \bigg) \bigg \{ \frac{3}{ \sqrt{ {3}^{2} + {4}^{2} } } \sin(5\pi t) + \frac{4}{ \sqrt{ {3}^{2} + {4}^{2} } } \cos(5\pi t) \bigg \}$

$= > x = 5 \bigg \{ \frac{3}{5} \sin(5\pi t) + \frac{4}{5} \cos(5\pi t) \bigg \}$

Let 3/5 be cos(θ) , then 4/5 will be sin(θ) .

$= > x = 5 \bigg \{ \sin(5\pi t) \cos( \theta) + \sin( \theta) \cos(5\pi t) \bigg \}$

$= > x = 5 \bigg \{ \sin(5\pi t + \theta) \bigg \}$

So maximum value of x is reached when value of sin component is 1

So max displacement will be 5 units.

Answer 2

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$\tt{\huge{\purple{\leadsto{ Question \::- }}}}$

The displacement of a particle is given by X equal to 3 sin 5πt the + 4 cos 5πt then the amplitude of particle is ..........

SOLUTION -

From the question, we can obtain the following Equation for the displacement of the particle -

Displacement Of Particle -: 3 Sin ( 5 π t ) + 4 cos ( 5 π t )

Let us find the Magnitude -

Magnitude - √[ { Coefficient of a } ^ 2 + { Coefficient of b } ^ 2 ]

Here,

a = 3

b = 4

Magnitude - √ [ ( 3 ) ^ 2 + ( 4 ) ^ 2 ] = 5

Now, =>

Multiplying and dividing -

=> 5 [ ( 3 / 5 ) × sin ( 5 π t ) + ( 4 / 5 ) × cos ( 5 π t ) ] = 0

=>5 [ ( 3 / 5 ) sin ( 5 π t ) + ( 4 / 5 ) cos ( 5 π t ) ] = 0

So, the maximum value is 5 as the maximum value of [ ( 3 / 5 ) sin ( 5 π t ) + ( 4 / 5 ) cos ( 5 π t ) ] = 1

So the amplitude of the particle is 5.

Answer :

The amplitude of the required particle is 5.