# Tutorial 1 ans

Tutorial 1 Question 1

• A heavy table is supported by flat steel legs. Its natural period in horizontal oscillation is 0.4s. When a 30‐kg plate is clamped to its surface, the natural period in the oscillation is increased to its surface the natural period in the oscillation is increased to 0.5s. What is the effective spring constant and the mass of the table?

• [Steidel, Problem 2.8, page 49]

MP3002/MP4012 Mechanics of

MP3002/MP4012 Mechanics of

Deformable Solids

Tutorial 1: Free Vibration

Tutorial 1: Free Vibration

by Force Method

1

2

Solution to Question 1

… Solution to Question 1

• This question shows the dependence of natural frequency on mass. Let us write equation of motion (EOM) for both cases mass Let us write equation of motion (EOM) for both cases

• Let us relate the measured natural period of oscillation to the derived theory for natural frequency.

derived theory for natural frequency

• Case (a) has the EOM being

mx kx 0

• Natural period of Case (a) is

– Obtained by force balance

• Case (b) has the EOM being

(m 30k ) k 0

30kg) x kx

1

m

1

m 30k

30kg

2

b 0.5s= 2

fa

k

fb

k

• C

Comparing the two measured natural period yields:

i th t

d t l

i d i ld

a

m 1

k

m

2

k 2 m 30k

30kg

m 30k

30kg

b

• Solving for the table mass m from the relationship above yields 2

0.82

2

a m 0 8 30kg=53.3kg

a

2

1 m 30kg

1 0.8

b

b

a 0.4s=

– Because of added 30‐kg mass

• Its natural frequency is

q

y

1 k

fa

2 m

• Natural period of Case (b) is

• Its natural frequency reduces to

Its natural frequency reduces to

1

k

fb

2 m 30kg

m

x

k

3

The table mass is calculated from the

added mass and ratio of period

4

… Solution to Question 1

Solution to Question 1

Tutorial 1: Question 2

Tutorial 1: Question 2

1

m

2

fa

k

• From the formula of period

of case (a)

a 0.4s=

• The stiffness is derived as

k m 2 f a

2

2

m

a

• A new model car is suspended as a pendulum, using inextensible cables attached to the front and rear axles. With the projected length of cable l=4.6m, the period of swinging the projected length of cable l=4 6m the period of swinging

oscillation is 4.3s. By shortening the length l=2.6m, the period decreases to 3.3s. Determine the distance h from the wheel axles to the car center of gravity.

• [Steidel, Problem 2.38, page 66]

2

• Substitution of the measured period a and calculated table mass, value of the stiffness is calculated as

l

f h

ff

l l d

m

2

k 30kg

13138.0 N/m

0.4s

2

x

k

5

6

… Solution to Q.2

… Solution to Q.2

Solution to Question 2

Solution to Question 2

• This question shows the dependence of natural frequency on q

p

q

y

moment of inertia, which is in turn related to the location of the center of gravity through the parallel axis theorem. • In this case, the car is suspended such that the pivot of cable suspension passes through the car center of gravity (CG) • As the car swings, the CG is perturbed off the initial vertical by an angular displacement θ

• As the distance between the CG and the pivot point is p

p

changed from l1h to l2h, the natural period of oscillation changed from 1 to 2.

• The moment of inertia about pivot changes as following

• Case 1

Case 1

J 1 J CG m l1 h

l1

l

CG

CG

h

1

CG

7

CG

• Case 2

Case 2

J 2 J CG m l2 h

2

h

l2

2

CG

CG

2

h

8

… Solution to Q. 2

Solution to Q 2

… Solution to Q. 2

Solution to Q 2

• The two cases differ in natural period

• Difference in moment of inertias between the two cases can be expressed in terms of natural periods

•...

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