A particle moves with S.H.M. in a straight line. In the first second a

1.	A particle moves with S.H.M. in a straight line. In the first second after starting from rest, it travels a distance x1 cm and in the next round it travels a distance x2 cm in the same direction. Prove that the amplitude of oscillation is \(\frac{2x^2_1}{3x_1-x_2}\).
Answer» As the particle starts from rest, it must start from the extreme position. Hence, when t = 0, x = r, where r is the required amplitude. Using the relation, x = rcos ωt Or r − x₁ = rcos ω × 1 = r cos ω …(i) And r − (x₁ + x₂ ) = rcos ω × 2 = rcos 2ω Or r − x₁ − x₂ = r(2cos²ω − 1) …(ii) Solving (i) and (ii), r = \(\frac{2x^2_1}{3x_1-x_2}\)

A particle moves with S.H.M. in a straight line. In the first second after starting from rest, it travels a distance x1 cm and in the next round it travels a distance x2 cm in the same direction. Prove that the amplitude of oscillation is \(\frac{2x^2_1}{3x_1-x_2}\).

Answer»

As the particle starts from rest, it must start from the extreme position. Hence, when t = 0, x = r, where r is the required amplitude.

Using the relation, x = rcos ωt

Or r − x₁ = rcos ω × 1

= r cos ω …(i)

And r − (x₁ + x₂ ) = rcos ω × 2

= rcos 2ω

Or r − x₁ − x₂ = r(2cos²ω − 1) …(ii)

Solving (i) and (ii),

r = \(\frac{2x^2_1}{3x_1-x_2}\)

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A particle moves with S.H.M. in a straight line. In the first second after starting from rest, it travels a distance x1 cm and in the next round it travels a distance x2 cm in the same direction. Prove that the amplitude of oscillation is \(\frac{2x^2_1}{3x_1-x_2}\).

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