Amplitude of an SHO is A. When it is at a distance y from the mean position of the path of its oscillation, the SHO receives blow in the direction of its motion, which doubles its velocity instantaneously. Find the new amplitude of its oscillations.

Amplitude of an SHO is A. When it is at a distance y from the mean pos

1.	Amplitude of an SHO is A. When it is at a distance y from the mean position of the path of its oscillation, the SHO receives blow in the direction of its motion, which doubles its velocity instantaneously. Find the new amplitude of its oscillations.
Answer» Solution :Velocity of SHO at displacement y. `v= omega_(0) sqrt(A^(2) - y^(2))"""……."(1)` Now instantaneous velocity after giving blow in the direction of motion is `v_(1)`, `v= omega_(0) sqrt(A^(2) - y^(2))` where `A_(1)` is new AMPLITUDE `2v= omega_(0) sqrt(A^(2) - y^(2))""[therefore v_(1)= 2v]` `therefore 2 omega_(0) sqrt(A^(2) - y^(2)) = omega_(0) sqrt(A_(1)^(2) - y^(2))""[=therefore " From EQ. (1) "]` `therefore 4 (A^(2)- y^(2))= A_(1)^(2) -y^(2)""[therefore " Squaring "]` `therefore 4A^(2) - 4Y^(2) = A_(1)^(2) -y^(2)` `therefore 4A^(2)- 3y^(2) = A_(1)^(2)` `therefore A_(1) = sqrt(4A^(2)- 3y^(2))`.