The resultant amplitude due to superposition of three simple harmonic motions x 1-3 sin- t - x 2-5 sin- t -37- and x 3-15 cos - t isA- -111B- -189C- 10D- 18

The correct option is B √(189)x_1=3 sin⁡ω t, x_2=5sin⁡(ω t+37^∘) x_3=15 sin⁡(ω t+270^∘ ) ∑ A_x=3+5 cos 37^∘ =3+5×4/5=7 ∑ A_4=5sin 37^∘ 15 =3 15= 12 A=√ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Global Maxima

The resultant...

Question

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 $x_{3} = - 15 c o s ω t$ is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{111}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{189}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\sqrt{189}$
$x_{1} = 3 s i n ω t, x_{2} = 5 s i n (ω t + 37^{\circ}) x_{3} = 15 s i n (ω t + 270^{\circ}) \sum A_{x} = 3 + 5 c o s 37^{\circ} = 3 + 5 \times \frac{4}{5} = 7 \sum A_{4} = 5 s i n 37^{\circ} - 15 = 3 - 15 = - 12 A = \sqrt{144 + 49} = \sqrt{189}$

Suggest Corrections

Similar questions

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. 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. The resultant amplitude due to superposition of two SHMs,

x_{1} = 10 s i n (ω t + 30^{\circ}

)cm and

x_{2}

= 10 cos (

ω t + 60^{\circ}

)cm is

Q. The resultant amplitude due to superposition of two waves

Y_{1} = 5 sin (ω t - k x)

and

Y_{2} = - 5 cos (ω t - k x - 150^{\circ})

Join BYJU'S Learning Program

The resultant amplitude due to superposition of three simple harmonic motions x1=3sin ωt,x2=5sin(ωt+37∘) and x3=−15cosωt is

The correct option is D √189x1=3sinωt,x2=5sin(ωt+37∘)x3=15sin(ωt+270∘)∑Ax=3+5cos 37∘=3+5×45=7∑A4=5sin 37∘−15=3−15=−12A=√144+49=√189

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 $x_{3} = - 15 c o s ω t$ is

The correct option is D $\sqrt{189}$
$x_{1} = 3 s i n ω t, x_{2} = 5 s i n (ω t + 37^{\circ}) x_{3} = 15 s i n (ω t + 270^{\circ}) \sum A_{x} = 3 + 5 c o s 37^{\circ} = 3 + 5 \times \frac{4}{5} = 7 \sum A_{4} = 5 s i n 37^{\circ} - 15 = 3 - 15 = - 12 A = \sqrt{144 + 49} = \sqrt{189}$