The total energy at any instant in a simple harmonic motion is constant and equal to ½m⍵²A². Hence the average energy in one time period will also be the same = ½m⍵²A². So option (a).
(b) an extreme position
EXPLANATION:
At the extreme position the particle is in the same phase at consecutive appearance, hence (b) is true. In all other options, the consecutive appearance of the...
The correct answer is
(d) zero.
EXPLANATION:
The displacement is a vector and its magnitude is given by the line joining the initial position to the final position. Since after one time period...
The correct answer is
(d) zero.
EXPLANATION:
Since the acceleration is proportional to the displacement and always directed towards the mean position, half of one time period it is directed towards left and...
(a) 1/2 m ω2 A2
EXPLANATION:
The total energy at any instant in a simple harmonic motion is constant and equal to ½m⍵²A². Hence the average energy in one time period will also...
The correct answer is
(c) 2 v
EXPLANATION:
Between one extreme position to another i.e. in half the time period the K.E. increases from zero to maximum (at the mean position) and again...
(d) the average potential energy in one time period equal to the average kinetic energy in this period.
EXPLANATION:
The sum of K.E. and P.E. is constant, they are not equal. (a)...
(a) the maximum potential energy equals the maximum kinetic energy
(b) the minimum potential energy equals the minimum kinetic energy
EXPLANATION:
Minimum P.E. and minimum K.E. are respectively at the mean position and at...
