Average value of kinetic energy and potential energy over entire time period in a SHM is

Question

Average value of kinetic energy and potential energy over entire time period in a SHM is (A) 0, (1/2) m ω2A2 (B)  (1/2) m ω2A2,0 (C)  (1/2) m ω2A2, (1/2) m ω2A2 (D)  (1/4) m ω2A2, (1/4) m ω2A2

Tardigrade Admin · Accepted Answer

Maximum KE =(1/2) m ω2 A2 Minimum K E=0 Average KE =((0+(1/2) m ω2 A2/2)) =(1/4) m ω2 A 2 Similarly average PE =((0+(1/2) m ω2 A2/2)) =(1/4) m ω2 A2

Q. Average value of kinetic energy and potential energy over entire time period in a SHM is

$0, \frac{1}{2} m ω^{2} A^{2}$

$\frac{1}{2} m ω^{2} A^{2}, 0$

$\frac{1}{2} m ω^{2} A^{2}, \frac{1}{2} m ω^{2} A^{2}$

$\frac{1}{4} m ω^{2} A^{2}, \frac{1}{4} m ω^{2} A^{2}$

Maximum $K E = \frac{1}{2} m ω^{2} A^{2}$
Minimum $K E = 0$
Average $K E = (\frac{0 + \frac{1}{2} m ω ^{2} A ^{2}}{2})$
$= \frac{1}{4} m ω^{2} A^{2}$
Similarly average $PE = (\frac{0 + \frac{1}{2} m ω ^{2} A ^{2}}{2})$
$= \frac{1}{4} m ω^{2} A^{2}$

Q. Average value of kinetic energy and potential energy over entire time period in a SHM is

0,21​mω2A2

21​mω2A2,0

21​mω2A2,21​mω2A2

41​mω2A2,41​mω2A2

Maximum KE=21​mω2A2 Minimum KE=0 Average KE=(20+21​mω2A2​) =41​mω2A2 Similarly average PE=(20+21​mω2A2​) =41​mω2A2

$0, \frac{1}{2} m ω^{2} A^{2}$

$\frac{1}{2} m ω^{2} A^{2}, 0$

$\frac{1}{2} m ω^{2} A^{2}, \frac{1}{2} m ω^{2} A^{2}$

$\frac{1}{4} m ω^{2} A^{2}, \frac{1}{4} m ω^{2} A^{2}$

Maximum $K E = \frac{1}{2} m ω^{2} A^{2}$
Minimum $K E = 0$
Average $K E = (\frac{0 + \frac{1}{2} m ω ^{2} A ^{2}}{2})$
$= \frac{1}{4} m ω^{2} A^{2}$
Similarly average $PE = (\frac{0 + \frac{1}{2} m ω ^{2} A ^{2}}{2})$
$= \frac{1}{4} m ω^{2} A^{2}$