Question

A 160 g block attached to a spring with spring constant 3.0 N/m oscillates horizontally on a frictionless table. Its velocity is 18 cm/s when x0 = -4.6 cm .

What is the amplitude of oscillation?

What is the block's maximum acceleration?

What is the block's position when the acceleration is maximum?

What is the speed of the block when x1x1x_1 = 3.2 cmcm ?

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Answer 1

Solu According to covestion, we have mass of block = 160g = 0.160kg = 2 Spring Constant = 3.0 N/m . = 18 cm3 = 0.18 ms when a = - 4.6cm = 0,046 from formula for velocity in stim, we have & V= wV A2 x²

9 + 0,002116 // 0.0324 18.75 mum 0.160 = 0.001728 +0.002116. = 0,0038 44 A = $0.008844 = 0.062 m. Demplitude of oscillation is 0.0624. 2) maximum acceleration The acceleration of block is maximum when it is at extreme position. i amax = w²A = (KJA * 0,062 = (18.75 0,062) m52 a max = 1.1625 m52 3) Position when acceleration is maximum, As we use in above question. The acceleration of block is maximum when it is at extezene Position il Position of block is its amplitude = 0,062m 0.160 a max
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