Question

A particle executes simple harmonic motion with an amplitude of 10cm if the angular frequency at oscillation of particle is 20rad/s the acceleration of the particle of it extreme position is

Answer 1

Answer:

• The acceleration of particle is 4000 cm/s²

Given:

1. Angular Frequency (ω) = 20 rad/sec
2. Amplitude (A) = 10 cm.

Explanation:

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From the formula we know,

⇒ a = ω² x

Substituting the values,

⇒ a = ω² × A

For Extreme position the condition should be x = A.

⇒ a = ω² × A

⇒ a = (20 rad/s)² × 10 cm

⇒ a = (20)² × 10

⇒ a = 400 × 10

⇒ a = 4000

⇒ a = 4000 cm/s²

∴ The acceleration of particle is 4000 cm/s².

Note:

• Symbols have their usual meanings.

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Extra Formulas:

⇒ v = ω √ ( A² - y² )

⇒ T = 2 π √ ( l / g )

⇒ T = 2 π √ ( m / k)

⇒ ω = √ ( K / m )

⇒ ω = √ ( a / x)

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Answer 2

Answer:

At an extreme position, the particle accelerates at 40m/s².

Explanation:

The particle's acceleration when performing simple harmonic motion is given as,

$a=\omega^2x$         (1)

Where,

a=acceleration of the particle

ω=angular velocity of the particle

x=displacement of the particle

From the question we have,

The angular velocity of the particle=20 rad/s

The amplitude=10cm=10×10⁻²m            (1cm=10⁻²m)

Since the particle is at an extreme position then its displacement will be maximum i.e.it will be equal to amplitude. So, equation (1) can be written as,

$a=\omega^2A$      (2)

By placing the values in equation (2) we get;

$a=(20)^2\times 10\times 10^-^2$

$a=400\times 0.1$

$a=40m/s^2$

Hence, the acceleration of the particle at an extreme position is 40m/s².

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