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Standard XII

Physics

Vector Addition

A particle in...

Question

A particle in SHM is described by the displacement

equation x(t) = A cos ( $ω t + θ)$ . If the initial (t = 0)

position of the particle is 1 cm and its initial

velocity is $π$ cm/s, what is its amplitude?

The angular frequency of the particle is $π s^{- 1}$

1 cm

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2 cm

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2.5 cm

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Solution

The correct option is
D

Given v = $π$ cm/sec , x = 1 cm and $ω = π s^{- 1}$

using v = $ω \sqrt{a^{2} - x^{2}} \Rightarrow π = π \sqrt{a^{2} - 1}$

$\Rightarrow 1 = a^{2} - 1 \Rightarrow a = \sqrt{2} c m$ .

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A particle in SHM is described by the displacement equation x(t) = A cos ( ωt+θ) . If the initial (t = 0) position of the particle is 1 cm and its initial velocity is π cm/s, what is its amplitude? The angular frequency of the particle is πs−1

The correct option is D Given v = π cm/sec , x = 1 cm and ω=πs−1 using v = ω√a2−x2⇒π=π√a2−1 ⇒1=a2−1⇒a=√2cm.

A particle in SHM is described by the displacement

equation x(t) = A cos ( $ω t + θ)$ . If the initial (t = 0)

position of the particle is 1 cm and its initial

velocity is $π$ cm/s, what is its amplitude?

The angular frequency of the particle is $π s^{- 1}$

The correct option is
D

Given v = $π$ cm/sec , x = 1 cm and $ω = π s^{- 1}$

using v = $ω \sqrt{a^{2} - x^{2}} \Rightarrow π = π \sqrt{a^{2} - 1}$

$\Rightarrow 1 = a^{2} - 1 \Rightarrow a = \sqrt{2} c m$ .