14.22 Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

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Vishal Baghel | Contributor-Level 10

5 months ago

The equation of displacement of a particle executing SHM at an instant t is given bas:

x = Asin $ω t$ , where A = Amplitude of oscillation and $ω$ = angular frequency = $\sqrt{\frac{k}{M}}$

The velocity of the particle, v = $\frac{d x}{d t}$ = A $ω \cos ? ω t$

The kinetic energy of the particle $E_{k}$ = $\frac{1}{2}$ M $v^{2}$ = $\frac{1}{2}$ M ${(A ω \cos ? ω t)}^{2}$ = $\frac{1}{2}$ M $A^{2} ω^{2} {c o s}^{2} ω t$

The potential energy of the particle $E_{p}$ = $\frac{1}{2}$ k $x^{2}$ = $\frac{1}{2}$ k $A^{2} {s i n}^{2} ω t$

For time period T, the average kinetic energy over a single cycle is given as :

${E_{k}}_{a v g} = \frac{1}{T} \int_{0}^{T} E_{k} d t$ = $\frac{1}{T} \int_{0}^{T} \frac{1}{2} M A^{2} ω^{2} {c o s}^{2} ω t d t$ = $\frac{M A^{2} ω^{2}}{2 T} \int_{0}^{T} {c o s}^{2} ω t d t$

= $\frac{M A^{2} ω^{2}}{2 T} \int_{0}^{T} \frac{(1 - c o s 2 ω t)}{2} d t$ = $\frac{M A^{2} ω^{2}}{2 T} ({t + \frac{s i n 2 ω t}{2 ω})}_{0}^{T}$ = $\frac{M A^{2} ω^{2}}{4 T} (T)$ = $\frac{1}{4} M A^{2} ω^{2}$ …….(i)

Average potenti

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The equation of displacement of a particle executing SHM at an instant t is given bas:

x = Asin $ω t$ , where A = Amplitude of oscillation and $ω$ = angular frequency = $\sqrt{\frac{k}{M}}$

The velocity of the particle, v = $\frac{d x}{d t}$ = A $ω \cos ? ω t$

The kinetic energy of the particle $E_{k}$ = $\frac{1}{2}$ M $v^{2}$ = $\frac{1}{2}$ M ${(A ω \cos ? ω t)}^{2}$ = $\frac{1}{2}$ M $A^{2} ω^{2} {c o s}^{2} ω t$

The potential energy of the particle $E_{p}$ = $\frac{1}{2}$ k $x^{2}$ = $\frac{1}{2}$ k $A^{2} {s i n}^{2} ω t$

For time period T, the average kinetic energy over a single cycle is given as :

${E_{k}}_{a v g} = \frac{1}{T} \int_{0}^{T} E_{k} d t$ = $\frac{1}{T} \int_{0}^{T} \frac{1}{2} M A^{2} ω^{2} {c o s}^{2} ω t d t$ = $\frac{M A^{2} ω^{2}}{2 T} \int_{0}^{T} {c o s}^{2} ω t d t$

Average potential energy over one cycle is given as :

${E_{p}}_{a v g} = \frac{1}{T} \int_{0}^{T} E_{p} d t$ = $\frac{1}{T} \int_{0}^{T} \frac{1}{2} M A^{2} ω^{2} \sin ? ω t d t$

= $\frac{M A^{2} ω^{2}}{2 T} \int_{0}^{T} \frac{(1 - c o s 2 ω t)}{2} d t$ = $\frac{M A^{2} ω^{2}}{2 T} ({t - \frac{s i n 2 ω t}{2 ω})}_{0}^{T}$ = $\frac{M A^{2} ω^{2}}{4 T} (T)$ = $\frac{1}{4} M A^{2} ω^{2}$ …….(ii)

From equations (i) and (ii), it is proved that average kinetic energy for a given period of time is equal to the average potential energy for the same period of time.

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14.22 Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

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