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A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm gong towards positive x-direction. Write the equation for the displacement of the particle at t = 4 s.

A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm gong towards positive x-direction. Write the equation for the displacement of the particle at t = 4 s.

Grade:11

2 Answers

Kevin Nash
askIITians Faculty 332 Points
9 years ago
Sol. Given, r = 10cm. At t = 0, x = 5 cm. T = 6 sec. So, w = 2π/T = 2π/6 = π/3 Sec-1 At, t = 0, x = 5 cm. So, 5 = 10 sin (w × 0 + ∅) = 10 sin ∅ [y = r sin wt] Sin ∅ = 1/2 ⇒ ∅ = π/6 ∴ Equation of displacement x = (10cm) sin (π/3) (ii) At t = 4 second X = 10 sin [π/3 × 4 × π/6] = 10 [8π+π/6] = 10 sin (3π/2) = 10 sin (π = π/2) = - 10 sin (π/2) = - 10 Acceleration a = – w2x = – (π2/9) × (- 10) = 10.9 = 0.11 cm/sec.
ankit singh
askIITians Faculty 614 Points
3 years ago
● Answer-
f = cosinv[(a+c)/2b] / 2πt0
● Explaination-
Let,
a = Asin(wt0)
b = Acos(2wt0)
c = Acos(3wt0)
Now take,
a + c = Asin(wt0) + Asin(3wt0)
a + c = 2Asin(2wt0).cos(wt0)
a + c = 2b.cos(wt0)
cos(wt0) = (a+c)/2b
Taking inverse on both sides,
wt0 = cosinv[(a+c)/2b]
2πft0 = cosinv[(a+c)/2b]
f = cosinv[(a+c)/2b] / 2πt0
Therefore, frequency of oscillation is cosinv[(a+c)/2b] / 2πt0 .

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