A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1 (t) = A sin ω t and x2(t) = Asin( ω t + ( 2 π / 3) ) Adding a third sinusoidal displacement x3(t)=B sin(ω l + Φ) brings the mass to a complete rest. The values of B and Φ are

Question

A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1 (t) = A sin ω t and x2(t) = Asin( ω t + ( 2 π / 3) ) Adding a third sinusoidal displacement x3(t)=B sin(ω l + Φ) brings the mass to a complete rest. The values of B and Φ are (A) √ 2 A , ( 3 π /4) (B)  A , ( 4 π /3) (C) √ 3 A , ( 5 π /6) (D)  A , ( π /3)

Tardigrade Admin · Accepted Answer

Resultant amplitude of x1 and x 2 is A at angle ((π/3)) from A1. To make resultant of x1 ,x2 and x3 to be zero. A3 should be equal to A at angle Φ =(4 π /3) as shown in figure. ∴ Correct answer is (b). Alternate Solution It we substitute, x1 +x2 + x3 =0 or Asin ω t + A sin ( ω t + ( 2 π/3) ) + B sin (ω t + Φ ) =0 Then by applying simple mathematics we can prove that B= A and Φ = ( 4 π/3)

Q. A point mass is subjected to two simultaneous sinusoidal
displacements in x-direction, $x_{1} (t) = A s inω t$ and
$x_{2} (t) = A s in (ω t + \frac{2 π}{3})$ Adding a third sinusoidal
displacement $x_{3} (t) = B s in (ω l + Φ)$ brings the mass to a
complete rest. The values of B and $Φ$ are

$2 A, \frac{3 π}{4}$

$A, \frac{4 π}{3}$

$3 A, \frac{5 π}{6}$

$A, \frac{π}{3}$

Q. A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1​(t)=Asinωt and x2​(t)=Asin(ωt+32π​) Adding a third sinusoidal displacement x3​(t)=Bsin(ωl+Φ) brings the mass to a complete rest. The values of B and Φ are

2​A,43π​

A,34π​

3​A,65π​

A,3π​

$2 A, \frac{3 π}{4}$

$A, \frac{4 π}{3}$

$3 A, \frac{5 π}{6}$

$A, \frac{π}{3}$