Question

Time period of a simple harmonic
motion is ts. If the velocity of the particle in
equilibrium position is 10 cm/s, then find the
length of oscillation (from one end to the other
end) and find the velocity of the particle at a
distance of 3 cm from the equilibrium.​

Answer 1

Answer:

Using the relation 

Using the relation

 v=w × route(A2−x2)   [v is the velocity as displacement x,  A :Amplitude]

  From given 

(i) when x=8,v=33=w route(A2−64 )........... (1)

........... (1)(ii) when x=6,v=4

........... (1)(ii) when x=6,v=44=w route(A2−36) .......... (2)

 .......... (2)dividing (1)÷(2) gives

 .......... (2)dividing (1)÷(2) gives(43)2=A2−36A2−64⇒A2=716×64−9×36=100

 .......... (2)dividing (1)÷(2) gives(43)2=A2−36A2−64⇒A2=716×64−9×36=100⇒A=10 cm

 .......... (2)dividing (1)÷(2) gives(43)2=A2−36A2−64⇒A2=716×64−9×36=100⇒A=10 cmPutting A in (1) gives

 .......... (2)dividing (1)÷(2) gives(43)2=A2−36A2−64⇒A2=716×64−9×36=100⇒A=10 cmPutting A in (1) gives3=w102−64⇒w=63=21 rad/s

 .......... (2)dividing (1)÷(2) gives(43)2=A2−36A2−64⇒A2=716×64−9×36=100⇒A=10 cmPutting A in (1) gives3=w102−64⇒w=63=21 rad/s∴ Time period ocsillaion 

 .......... (2)dividing (1)÷(2) gives(43)2=A2−36A2−64⇒A2=716×64−9×36=100⇒A=10 cmPutting A in (1) gives3=w102−64⇒w=63=21 rad/s∴ Time period ocsillaion T=w

I HOPE ITS HELPFUL TO YOU.