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A body executes SHM on a line of length 16 cm with a frequency 1/24 find average velocity when it moves from mean position to half of amplitude???

Indraja , 6 Years ago

Grade 12

2 Answers

Eshan

Last Activity: 6 Years ago

Dear student,

Let the SHM be represented as $x=A sin\omega t$
$\implies v=A\omega cos\omega t$

Hence average velocity over $t=0$ to $t=\dfrac{\pi}{6\omega}$ is
$\dfrac{\int_0^{\pi/6\omega}A\omega cos\omega t}{\pi/6\omega}$

$=\dfrac{3A\omega}{\pi}$

$\omega=2\pi f=\dfrac{\pi}{12}$
$\implies v_{avg}=4cm/s$

Khimraj

Last Activity: 6 Years ago

time taken by body to reach half of amplitude from mean position T’ = T/12

v = Acoswt

v_avg = (1/T’) $\int_{0}^{T`} Acos(wt)dt$

v_avg = (12A/Tw) $[sinwt]_{0}^{T/12}$

v_avg = (6A/ $\pi$ )sin $\pi$ /6

v_avg = 3A/ $\pi$

here A = 8cm

SO v_avg = 24/ $\pi$ cm/s.

Hope it clears.

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