(a) Explanation Given Information: The mass ( m ) of the block is 50 kg. The amplitude ( x m ) of the resulting motion is 60 mm. The spring constant ( k 1 ) of top spring is 24 kN/m . The spring constant ( k 2 ) of bottom spring at left is 12 kN/m . The spring constant ( k 3 ) of bottom spring at right is 12 kN/m . Calculation: Show the free body diagram of the spring arrangement as in Figure (1): Springs are connected in parallel, thus the deflection in each spring is equal. δ 1 = δ 2 = δ 3 = δ Springs are connected in parallel, thus the load on the entire spring mass system is the sum of loads in each spring. Express the load on the entire spring mass system: P = P 1 + P 2 + P 3 (1) Here, the load on the entire spring mass system is P and loads on individual springs are P 1 , P 2 and P 3 . The load on a spring is the product of its spring constant and the deflection. Rewrite the Equation (1) for total load: P = P 1 + P 2 + P 3 k δ = k 1 δ 1 + k 2 δ 2 + k 3 δ 3 (2) Substitute δ 1 = δ 2 = δ 3 = δ in Equation (2). k δ = k 1 δ + k 2 δ + k 3 δ k = k 1 + k 2 + k 3 Substitute 24 kN/m for k 1 , and 12 kN/m for k 2 and k 3 (b) To determine The maximum velocity ( v m ) and acceleration ( a m ) of the block.

12th Edition

ISBN: 9781264095032

Author: BEER

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 19.1, Problem 19.21P

(a)

To determine

The period (τn) and frequency (fn) of the motion.

(b)

To determine

The maximum velocity (vm) and acceleration (am) of the block.

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Chapter 19.1, Problem 19.20P

Chapter 19.1, Problem 19.22P

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The period ( τ n ) and frequency ( f n ) of the motion.

Solution Summary: The author shows the free body diagram of the spring arrangement as in Figure (1): Springs are connected in parallel, thus the deflection in each spring is equal.

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The period (τn) and frequency (fn) of the motion.

The maximum velocity (vm) and acceleration (am) of the block.

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Chapter 19 Solutions