Explanation Given information: The mass of the block A ( m A ) is 2.4 kg. The mass of the block B ( m B ) is 0.9 kg. The spring constant ( k ) is 180 N/m. The coefficient of damping ( c ) is 60 N ⋅ s / m . The time ( t ) is 0.1 sec. The acceleration due to gravity ( g ) is 9 .81m/s 2 . Calculation: Calculate the natural circular frequency ( ω n ) using the formula: ω n = 2 k m A Here, 2 k is the spring constant of two springs. Substitute 180 N/m for k and 2.4 kg for m A . ω n = 2 × 180 N/m 2.4 = 150 = 12.24745 rad/s Calculate the critical damping coefficient ( c c ) using the formula: c c = 2 m A ω n Substitute 2.4 kg for m A and 12 .24745rad/s for ω n . c c = 2 × 2.4 × 12.24745 = 58.7878 N ⋅ s / m Calculate the stretch at equilibrium using the relation: Stretch at equilibrium = m A g 2 k Substitute 2.4 kg for m A , 9 .81m/s 2 for g , and 180 N/m for k . Stretch at equilibrium = 2.4 × 9.81 2 × 180 = 0.0654 m Calculate the initial stretch using the relation: Initial Stretch = ( m A + m B ) g 2 k Substitute 2.4 kg for m A , 0.9 kg for m B , 9 .81m/s 2 for g , and 180 N/m for k . Initial Stretch = ( 2.4 + 0.9 ) × 9.81 2 × 180 = 0.089925 m Since the coefficients c > c c which is critically over damped. Calculate the initial displacement ( x ) using the relation: x = Initial stretch- Stretch at equilibrium Substitute 0.089925 m for Initial stretch and 0.0654 m for Stretch at equilibrium. x = 0.089925 − 0.0654 = 0.024525 m The expression of the differential equation for the critically over damped as follows: x = c 1 e λ 1 t + c 2 e λ 2 t (1) Calculate the wavelength ( λ ) using the relation: m λ 2 + c λ + 2 k = 0 Here, 2 k is the spring constant of two springs. Substitute 2.4 kg for m , 60 N ⋅ s / m for c , and 180 N/m for k . 2.4 λ 2 + 60 λ + 2 ( 180 ) = 0 2.4 λ 2 + 60 λ + 360 = 0 Solve the above equation. λ = − b ± b 2 − 4 a c 2 a Substitute 60 for b , 2

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Author: BEER

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Chapter 19.5, Problem 19.138P

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Find the velocity of block A (v) after 0.1 sec.

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Chapter 19.5, Problem 19.137P

Chapter 19.5, Problem 19.139P

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Find the velocity of block A ( v ) after 0.1 sec.

Solution Summary: The author calculates the velocity of block A (v) after 0.1 sec.

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Find the velocity of block A (v) after 0.1 sec.

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