A mass of 0.26 kg on the end of a spring oscillates with a period of 0.45 s and an amplitude of 0.15 m .a) Find the velocity whe
n it passes the equilibrium point.b) Find the total energy of the system.c) Find the spring constant.d) Find the maximum acceleration of the mass.
       
      
                
     
    
    
    
    
    1 answer:
            
              
              
                
                
Answer:
a)V= 2.09 m/s
b)TE= 0.56 J
c)K=50.47 N/m
d)a=29.23 m/s²
Explanation:
Given that
m = 0.26 kg  ,  T= 0.45 s  ,A= 0.15 m
We know that time period given as

ω =Angular frequency


ω = 13.96 rad/s
The velocity at equilibrium
V=  ω A
V= 13.96 x 0.15 
V= 2.09 m/s
The total energy TE 


TE= 0.56 J
The spring constant K
Maximum stored energy in the spring 

From energy balance
U= TE

K A² = m V²


K=50.47 N/m
The maximum acceleration a
a= ω² A
a = 13.96²  x 0.15 m/s²
a=29.23 m/s²
  
                                
             
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