Tutorial 1 ans

Tutorial 1 ‐ Question 1
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

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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

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The table mass is calculated from the added mass and ratio of period

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… Solution to Question 1
Solution to Question 1

Tutorial 1: Question 2