Question

The SHM of a particle is given by x(t)=5 cos(2πt+π/4) in MKS units.Calculate the displacement and the magnitude of acceleration of the particle at t=1.5 s.

Answer 1

Given,

The simple harmonic motion of a particle is given by, x(t) = 5cos(2πt + π/4) in MKS units.

To find,

displacement and magnitude of acceleration of the particle at t = 1.5 sec.

displacement of particle, x(1.5) = 5cos(2π × 1.5 + π/4) = 5cos(3π + π/4) = -5cos(π/4) = -5/√2 m

acceleration of particle, a = d²x(t)/dt² = -20π²cos(2πt + π/4)

= -20π²cos(3π + π/4) = 20π²cos(π/4) = 10√2π²

Therefore displacement of particle is -5√2 m and acceleration of particle is 10√2π² m/s².

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Answer 2

Answer:

• Displacement (x) of the particle is -2.5 √{2} m
• Acceleration (a) of the particle is 10 √{2} π² m/s²

Given:

1. Given Equation:- x (t) = 5 cos (2 π t + π / 4)
2. Time period (t) = 1.5 seconds

Explanation:

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From given equation,

⇒ x (t) = 5 cos (2 π t + π /4)

Substituting the value of time (t) = 1.5 secs

⇒ x (1.5) = 5 cos (2 π × {1.5} + π /4)

⇒ x = 5 cos (3 π + π / 4)

As we know that the angle (3 π + π / 4) will lie in the third quadrant, where only tan functions are positive.

Therefore,

⇒ cos (3 π + θ) = - sin θ

Substituting in the above equation,

⇒ x = 5 × - sin(π / 4)

⇒ x = - 5 sin (π / 4)

⇒ x = - 5 × 1 / √{2}      ∵ [sin(π / 4) =1 / √{2}]

⇒ x = - 5 / √{2}

Rationalizing gives,

⇒ x = (- 5 × √{2}) / (√{2} × √{2})

⇒ x = (- 5 √{2}) / 2

⇒ x = - 2.5 √{2}

⇒ x = - 2.5 √{2} m

∴ Displacement (x) of the particle is -2.5 √{2} m.

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From the given equation we can make out,

⇒ x (t) = 5 cos (2 π t + π /4)

• Amplitude (A) = 5 m
• Angular Frequency (ω) = 2 π
• Phase (∅) = π / 4

Now, as we know standard acceleration formula,

⇒ a = - A ω² sin(ω t ± ∅)

Substituting the values,

⇒ a = - 5 (2 π)² sin (2 π t + π / 4)

⇒ a = - 5 × 4 π² × sin (2 π t + π /4)

We know that the angle (3 π + π / 4) will lie in the third quadrant, where only tan functions are positive.

Therefore,

⇒ sin (3 π + θ) = - cos θ

Substituting in the above equation,

⇒ a = - 5 × 4 π² × - cos (π /4)

⇒ a = -20 π² × - 1 / √{2}      ∵ [sin(π / 4) =1 / √{2}]

⇒ a = 20 π² × 1 / √{2}

⇒ a = 20 π² / √{2}

⇒ a = 10 √{2} π²

⇒ a = 10 √{2} π² m/s²

∴ Acceleration (a) of the particle is 10 √{2} π² m/s².

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