Explanation Given: Mass of bob is 1 , 2 kg . Length of pendulum is 600 mm . Mass of collar is 1.4 kg . Force acted upon is P m sin ω f t . The value of P m is 0.5 N . The value of ω f is 3 rad/s . θ is the small angle. Concept used: Draw the diagram for system as shown below: As the bob has a small displacement, from the above figure, θ = ( x − X ) l Draw the FBD for collar and the bob as shown below: Write the expression for natural frequency of the system. ω n = g l ...... (1) Here, ω n is the natural frequency of system, g is the acceleration due to gravity and l is the length of pendulum. Write the equation of motion for the bob. − F sin θ = m d 2 x d t 2 For small value of θ the value of sin θ ≈ θ . Substitute θ for sin θ in the above expression. − F θ = m d 2 x d t 2 Substitute m g for F in the above expression. − m g θ = m d 2 x d t 2 − g θ = d 2 x d t 2 Substitute ( x − X ) l for θ in the above expression. − g ( x − X ) l = d 2 x d t 2 Substitute ω n 2 for g l in above expression and simplify. d 2 x d t 2 + ω n 2 x = ω n 2 X ...... (2) Here, X is the initial displacement of collar. Write the expression for the initial displacement of collar as. X = X m sin ω f t Here, X m is the magnitude of harmonic displacement, ω f is the frequency of motion and t is time period. Solve equation (2) for the value of x . x = x m sin ( ω f t ) ...... (3) Here, x m is the amplitude of deflection of bob. Write the expression for the amplitude of deflection of bob. x m = ω n 2 X m ( ω n 2 − ω f 2 ) ...... (4) Write the equation of motion for collar. P + F θ = M d 2 X d t 2 Substitute m g for F and ( x − X ) l for θ in the above expression. P + m g ( x − X ) l = M d 2 X d t 2 Substitute ω n 2 for g l in above expression and simplify. M d 2 X d t 2 + m ω n 2 X = P + m ω n 2 x Substitute P m sin ( ω f t ) for P and x m sin ( ω f t ) for x in above expression. M d 2 X d t 2 + m ω n 2 X = ( P m + m ω n 2 x m ) sin ( ω f t ) ...... (5) Here, x m is the amplitude of deflection of bob and P m is the amplitude of applied harmonic force. Here, X m is the amplitude of displacement of collar. Divide equation (4) by M . d 2 X d t 2 + m ω n 2 X M = ( P m + m ω n 2 x m ) M sin ( ω f t )

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ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Chapter 19.4, Problem 19.103P

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Chapter 19 Solutions

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Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - Prob. 19.3P Ch. 19.1 - Prob. 19.4P Ch. 19.1 - Prob. 19.5P Ch. 19.1 - A 20-lb block is initially held so that the...Ch. 19.1 - Prob. 19.7P Ch. 19.1 - A simple pendulum consisting of a bob attached to...Ch. 19.1 - Prob. 19.9P Ch. 19.1 - A 5-kg fragile glass vase is surrounded by packing...

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