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Two light springs of force constants k1 and k2 each 3.2 Nm-1 and a block of mass 200g are in one line AB on a smooth horizontal table such that one end of the each spring is fixed on a rigid support and the other end is free as shown fig. The distance CD between the free ends of a spring is 60cm. If block moves along AB with a velocity 120 cm/s in between the springs, the period of oscillations of the block is ?

Hrishant Goswami , 11 Years ago
Grade 10
anser 1 Answers
Vikash Tiwari

Last Activity: 6 Years ago

Let us consider body will start oscillation from point C ,b/w C and D.
First calculate time b/w C and D 
V=120cm/sec        displacement= 60cm 
Therefore time(CD)=60/120=1/2
Same as above time (DC)=1/2
Now
Total Time Period
T=T/2+t(CD)+T/2+t(DC)
       
        T=2π√m/k1/2 . +1/2 . +2π√m/k2/2 +1/2
T=2π(√m/k1+√m/k2)/2+1/2+1/2
 
T=π{√m/k1+√m/k2}+1
 
 
T=3.14(√0.2/1.8+√0.2/3.2)+1
T=3.14(√1/9+√1/16)+1
T=3.14(1/3+1/4)+1
T=3.14×7/12+1
T=1.83+1
T=2.83
T
 

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