# ME3112 - Free Vibration - Solution

January 11, 2018 | Author: Leonard | Category: Oscillation, Theoretical Physics, Mechanics, Classical Mechanics, Physics

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Question 1 Assume small oscillation, the cylinder rotates about C. Based on this, the following relations can be defined π₯π = ππ, and π₯π΄ = 2ππ where π₯π and π₯π΄ are the cylinder displacement at point O and A from its equilibrium position. And at equilibrium position, the moment equation about the contact point gives ππΏ2π β ππππ ππ(15π ) = 0 The spring force at the positive displaced position is πΉπ = π(π₯π΄ + πΏ) = 2πππ + ππΏ Now, from FBD of the cylinder

ο± πΌπΜ o

15

C

C mg

Taking moment about C gives the equation of motion of the system 2ππΉπ β ππππ ππ(15π ) = πΌπΜ + πππ₯Μ π βΉ 2π(2πππ + ππΏ) β ππππ ππ(15π ) = πΌπΜ + πππ₯Μ π βΉ 4π 2 ππ + [ππΏ2π β ππππ ππ(15π )] = πΌπΜ + πππ₯Μ π π₯π 1 π₯Μ π βΉ 4π 2 π ( ) = ( ππ 2 ) ( ) + πππ₯Μ π , π 2 π 3 βΉ ππ₯Μ π + (4π)π₯π = 0, or 2 8π βΉ π₯Μ π + ( ) π₯ = 0, ππ 3π π

1 or 4π 2 ππ = ( ππ 2 ) πΜ + ππ 2 πΜ 2 3 ππΜ + (4π)π = 0 2 8π πΜ + ( )π₯ = 0 3π π

Hence, the natural frequency of this system is ππ = β(

8π 8 200 )=β ( ) = 7.303 πππ/π  3π 3 10

(a) And its period is ππ =

2π 2π = = π. πππ π  ππ 7.303

(b) Given that it is displaced by 5 cm, and then set into vibration, which is given by π₯π = 0.05cos(ππ π‘) βΉ π₯Μ π = β0.05ππ 2 cos(π‘) Hence, the maximum acceleration during the vibration is (π₯Μ π )πππ₯ = 0.05ππ 2 = 0.05(7.303)2 = π. ππ ππ  β2

*Note: In this example, you can again choose to ignore the effects of the initial spring force and the weight of the cylinder (because they counter-balanced each other during the oscillation). This condition, however, cannot be generalized to all cases, such as in Question 2. So if you are not sure whether this condition is satisfied or not, please follow the above approach.

Question 2 At equilibrium, the FBD is

πΏ ππ ( π πππΌ1 ) = ππΏ0 πΏπππ πΌ3 2

Now, suppose the rod oscillates by a small angular displacement π, the FBD becomes

Again, taking moment about O, the EOM of the rod is πΏ 1 πΏ 2 βππ ( π ππ(πΌ1 + π)) + (ππΏ0 πππ πΌ3 β ππΏπ)πΏ = ( ππΏ2 + π ( ) ) πΜ 2 12 2 πΏ 1 β βππ (π πππΌ1 πππ π + πππ πΌ1 π πππ) + (ππΏ0 πΏπππ πΌ3 β ππΏ2 π) = ππΏ2 πΜ 2 3 Using small angle assumption, πΏ 1 β βππ (π πππΌ1 + πππ πΌ1 π) + (ππΏ0 πΏπππ πΌ3 β ππΏ2 π) = ππΏ2 πΜ 2 3 1 πΏ πΏ β ππΏ2 πΜ + [ππΏ2 + ππ πππ πΌ1 ] π + [ππ π πππΌ1 β ππΏ0 πΏπππ πΌ3 ] = 0 3 2 2 1 πΏ β ππΏ2 πΜ + [ππΏ2 + ππ πππ πΌ1 ] π = 0 3 2 Hence, the natural angular frequency is πΏ ππΏ2 + ππ 2 πππ πΌ1 ππ = β 1 2 3 ππΏ

Question 3 Position 1: Suppose collar A is offset from its equilibrium position by xA . At this position, the system has maximum potential energy (due to spring), but zero kinetic energy. Position 2: And during oscillation, the system will attain zero potential energy when collar A is at the equilibrium position, which is also when it has the maximum kinetic energy. At position 1, the following displacement diagrams can be drawn

ο’

From the geometry of the system, one observes that π₯π΄ = πππππ π½ = (ππππ π½)π And for SHM, π = π©π ππ(ππ π‘ + π) β (π₯π΄ )πππ₯ = (ππππ π½)π© Hence, the elastic potential energy is 1 1 2 π1 = π(π₯π΄ )πππ₯ = π(ππππ π½)2 π©2 2 2 At position 2, the velocity diagram at that instant can be drawn as shown.

Consider the velocities of collars A and B, π£βπ΅ = π£βπ΄ + π£βπ΅/π΄ β βπ£π΅ πβ = π£π΄ πβ + πΜ πββ Γ πβπ΅/π΄ β π£π΅ πβ = π£π΄ πβ + πΜ πββ Γ (βππ πππ½πβ + ππππ π½πβ) And matching the velocity components, give β π£π΄ = ππππ π½πΜ,

and

π£π΅ = ππ πππ½πΜ

Similarly, the velocity at G is π£βπΊ = π£π΄ πβ + πΜ πββ Γ πβπΊ/π΄ π π = (ππππ π½πΜ )πβ + πΜ πββ Γ (β π πππ½πβ + πππ π½πβ) 2 2 π π = ( πππ π½πΜ ) πβ β ( π πππ½πΜ ) πβ 2 2 Hence, the kinetic energy at this instant is 1 1 1 1 π2 = π|π£βπΊ |2 + πΌπΊ πΜ 2 + ππ |π£βπ΄ |2 + ππ |π£βπ΅ |2 2 2 2 2 1 π 2 1 1 1 1 2 2 = π ( πΜ ) + ( ππ 2 ) πΜ 2 + ππ (ππππ π½πΜ ) + ππ (ππ πππ½πΜ ) 2 2 2 12 2 2 1 1 = ( π + ππ ) π 2 πΜ 2 2 3 1 1 2 3

= ( π + ππ ) π 2 (ππ π©)2

(since π = π©π ππ(ππ π‘ + π))

Finally, applying the principle of conservation of energy, gives π1 = π2 1 1 1 β π(ππππ π½)2 π©2 = ( π + ππ ) π 2 (ππ π©)2 2 2 3 πππππ π· β ππ = β π (π π + ππ )

Question 4 The FBD of this system when it is subjected to small oscillation is

Taking moment about A, we get

And for small angle rotation, this is reduced to

Substituting the numeric, gives

Hence, the EOM for this system when expressed in the equivalent 1D spring-mass-damper system is

And its critical damping would be given by

And hence the damping ratio is π=

πππ 0.09 = = π. ππππ πππ 2.087