Three particles execute SHM along x- direction with same frequency as x1 - 5 sin- t- x2 - 5 sin - t - 53- x3 - -10 cos- t- Find the amplitude of the resultant SHM-

Given, x_1 = 5 sinω t .....(1) x_2 = 5 sin (ω t + 53^∘) ......(2) x_3 = 10 cosω t ......(3) We can write (3) as , x_3 = 10 sin (ω t + 270^∘) Finding t ...

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Standard XIII

Physics

SHM expression

Three particl...

Question

Three particles execute SHM along $x -$ direction with same frequency as $x_{1} = 5 sin ω t$ , $x_{2} = 5 sin (ω t + 53^{\circ})$ , $x_{3} = - 10 cos ω t$ . Find the amplitude of the resultant SHM.

6 units

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8 units

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10 units

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15 units

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Solution

The correct option is C $10 units$
Given,
$x_{1} = 5 sin ω t . . . . . (1)$
$x_{2} = 5 sin (ω t + 53^{\circ}) . . . . . . (2)$
$x_{3} = - 10 cos ω t . . . . . . (3)$
We can write $(3)$ as , $x_{3} = 10 sin (ω t + 270^{\circ})$
Finding the resultant amplitude using vector notation.

Resultant amplitude, $| R | = \sqrt{8^{2} + 6^{2}} = 10 units$

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Similar questions

Q. Three particles execute SHM along

x -

direction with same frequency as

x_{1} = 5 sin ω t

x_{2} = 5 sin (ω t + 53^{\circ})

x_{3} = - 10 cos ω t

. Find the amplitude of the resultant SHM.

Q. A particle is subjected to two SHMs of displacement

x_{1} = 5 sin (ω t + 30^{\circ})

and

x_{2} = 3 sin (ω t + 120^{\circ})

respectively. Find out the amplitude of the resulting SHM.

Q. A particle is subjected to two SHMs of displacement

x_{1} = 5 sin (ω t)

and

x_{2} = 3 sin (ω t + 60^{\circ})

respectively. Find out the equation of displacement of the resulting SHM.

Q. The resultant amplitude due to superposition of three simple harmonic motions

x_{1} = 3 sin ω t

x_{2} = 5 sin (ω t + 37^{\circ})

and

x_{3} = - 15 cos ω t

is:

Q. The resultant amplitude due to superposition of three simple harmonic motions

x_{1} = 3 s i n ω t, x_{2} = 5 s i n (ω t + 37^{\circ})

and

x_{3} = - 15 c o s ω t

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SHM expression

Standard XIII Physics

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Three particles execute SHM along x− direction with same frequency as x1=5sinωt, x2=5sin(ωt+53∘), x3=−10cosωt. Find the amplitude of the resultant SHM.

The correct option is C 10 unitsGiven, x1=5sinωt .....(1) x2=5sin(ωt+53∘) ......(2) x3=−10cosωt ......(3) We can write (3) as , x3=10sin(ωt+270∘) Finding the resultant amplitude using vector notation. Resultant amplitude, |R|=√82+62=10 units

Three particles execute SHM along $x -$ direction with same frequency as $x_{1} = 5 sin ω t$ , $x_{2} = 5 sin (ω t + 53^{\circ})$ , $x_{3} = - 10 cos ω t$ . Find the amplitude of the resultant SHM.