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# A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

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1.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

l+x m Figure 6.1 by, say, wrapping the spring around a rigid massless rod). The equilibrium length of the spring is ‘. Let the spring have length ‘ + x(t), and let its angle with the vertical be µ(t). Assuming that the motion takes place in a vertical plane, ﬂnd the equations of motion for x and µ.

2.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

This makes a pendulum of 2.8 m length swing through a distance 12.5 cm horizontal. Calculate metacentric height of the vessel. [6] … Mass of the connecting rod is 40 kg. [6] Total No. of Questions12] … A cantilever of 3 m length and uniform rectangular cross-section 150 mm wide and 300 mm deep is loaded with 30 kN load at its free end.

3.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

A particle of mass m slides back and forth between two inclined frictionless planes as shown. … A loaded vertical spring executes simple harmonic oscillations with period of 4 s. … A uniform rod of length and mass m is hinged at its lowest point O A and is connected at its highest point A by means of a spring of spring constant k as shown …

4.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

(b) A pendulum is formed by pivoting a long thin rod of length L and mass m about a pivot on the rod that is that is distance d above the center of rod, (i) find small amplitude period of this pendulum in terms of d, L, m, and g (ii) Show that the period has a minimum value when d =0.289L. Q # 7. (a) What is Doppler effect?

5.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

ii. The velocity of ball 2 and label the graph “2.”. iii. The velocity of the center of mass of the two-ball system and label the graph “C.”. potential energy as zero on the ground. At time t = 0, E = 3mgH , where m is the mass of one ball. and H is the height of the building. At time t = 2T , E = 2mgH .

6.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

Its speed is v and its distance from the center of the earth is R. Draw a free body diagram of the satellite. Draw another that takes account of the slight drag force of the earths atmosphere on the satellite. v* m R. problem 3.8: (Filename:ch2.6) 3.9 A uniform rod of mass m rests in the back of a flatbed truck as shown in the figure.

7.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

8.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

3.9 The uniform rigid rod shown in the gure hangs in the vertical plane with the support of the spring shown. In this position the spring is stretched from its rest length. Draw a free body diagram of the spring. Draw a free body diagram of the rod. x x m = 10 kg y frictionless problem 3.12: F = 50 N /3 m k. 9a A. 5a C 6a (Filename:pgure2.1 …

9.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

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10.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

To understand these factors, think of a pendulum. Pendulums move much more slowly than sound vibrations, but their principles are the same. Displace a pendulum to one side, and a restoring force (gravity) causes it to swing back toward its center rest position. But its inertia causes it to overshoot, leading to displacement in the other direction.

11.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

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12.A uniform rod of mass m and length l is free to swing back and forth by pivoting a distance x from its center. it undergoes harmonic oscillations by swinging back and forth under the influence of gravity.

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