A point mass is subjected to two simultaneous sinusoidal displacements in x-direction x1t-A sin- t and x2t- Asin - t-2 -3- Adding a third sinusoidal displacement x 3 t - B sin - t - brings the mass to a complete rest- The values of B and - areA- -2 A - 3 -4B- A- 4 -3C- A- -2D- -3 A - 5 -6

The correct option is B A, 4 π/3 As per the question, A sin ω t + A sin (ω t + 2 π/3) + B sin (ω t + ϕ) = 0 ⇒ B sin (ω t + ϕ) = [ A sin ω t + A si ...

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Byju's Answer

Standard XII

Physics

SHM expression

A point mass ...

Question

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

$\sqrt{2} A, \frac{3 π}{4}$

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$A, \frac{4 π}{3}$

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$\sqrt{3} A, \frac{5 π}{6}$

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$A, \frac{π}{3}$

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Solution

The correct option is B
$A, \frac{4 π}{3}$

As per the question, $A s i n ω t + A s i n (ω t + \frac{2 π}{3}) + B s i n (ω t + ϕ) = 0 \Rightarrow B s i n (ω t + ϕ) = - [A s i n ω t + A s i n (ω t + \frac{2 π}{3})] = - A [2 s i n (ω t + \frac{π}{3}) c o s \frac{π}{3}] \Rightarrow B s i n (ω t + ϕ) = - A s i n (ω t + \frac{π}{3}) = A s i n (ω t + \frac{4 π}{3})$
Hence, $B = A$ and $ϕ = \frac{4 π}{3}$

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