A force F = 10 sin at N acts on a displacement of x = 2 sin (t – a16)m. Determine (a) the work done during the first 6 s; (b) the work done during the first 12s. Solution: Rewriting Eq. (3.7-1) as W=[Fx'dt and substituting F = Fo sin t and x = X sin ( I- ) gives the work done per cycle of W=nF0X0 sin o For the force and displacement given in this problem, Fo = 10 N, X = 2 m, = 7/6and the period = 2s. Thus, in the 6 s specified in (a), three complete cycles take place, and the work done is W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s. W=WF0X0[cos 30°S01/2sin nt cos nt dt+sin 30°J01/2sin nt dt] =nx10x2[-0.8664rcos 2nt+0.50(12-sin 2rt47)]01/2 =16.51N:1

A force F = 10 sin at N acts on a displacement of x = 2 sin (t – a16)m. Determine (a) the work done during the first 6 s; (b) the work done during the first 12s. Solution: Rewriting Eq. (3.7-1) as W=[Fx'dt and substituting F = Fo sin t and x = X sin ( I- ) gives the work done per cycle of W=nF0X0 sin o For the force and displacement given in this problem, Fo = 10 N, X = 2 m, = 7/6and the period = 2s. Thus, in the 6 s specified in (a), three complete cycles take place, and the work done is W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s. W=WF0X0[cos 30°S01/2sin nt cos nt dt+sin 30°J01/2sin nt dt] =nx10x2[-0.8664rcos 2nt+0.50(12-sin 2rt47)]01/2 =16.51N:1

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Free undamped vibrations

Whenever an object/system displaces in either side of its equilibrium position with the help of an external/internal means, then the object/system moves regularly or randomly on both sides of its equilibrium position. This motion is called vibration. Generally, there are three types of vibration: longitudinal vibration, transverse vibration, and torsional vibration.

Question

plz do this problem unable to understand explain all steps related to vibration

A force F = 10 sin at N acts on a displacement of x = 2 sin (nt – n16)m. Determine (a) the work done during the first 6 s; (b) the
work done during the first 12s.
Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sin
t and x = X sin (
t- ) gives the work done per cycle of
W=nF0X0 sin ở
For the force and displacement given in this problem, Fo = 10 N, X = 2 m,
= n/6and the period
= 2s. Thus, in the 6 s specified in (a),
three complete cycles take place, and the work done is
W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m
The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s.
W=WF0X0[cos 30°f01/2sin nt cos at di+sin 30°f01/2sin nt dt]
=Tx10x2[-0.8664rcos 2rt+0.50(12–sin 2rt47)]01/2
=16.51N:1

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