A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is 
cm. The value of x is__________.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(700)
A = 10 cm
Total Energy = 
KA 2
By energy conservation we can final v at x = 5
Now, velocity is tripled through external mean so the amplitude of SHM will charge and so the total energy, (but potential) energy at this moment will remain same)
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
Two massless springs with spring constants 2 k and 2 k, carry 50 g and 100 g masses at their free e…
The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is e…
The equation of a particle executing simple harmonic motion is given by . At t=1s, the speed of pa…
The motion of a simple pendulum excuting S.H.M. is represented by following equation.
, where time…
A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6s. The time i…
The length of a seconds pendulum at a height h = 2R from earth surface will be: (Given: R=Radius of…