The maximum velocity of a particle- executing simple harmonic motion with an amplitude 7 mm- is 4-4 m-s- The period of oscillation is 0-1 s0-01 s100 s10 s

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Toppr Admin · Accepted Answer

For a SHM- x-Asin-x3C9-t-x3D5-Velocity- v-dxdt-A-x3C9-cos-x3C9-t-x3D5-So- vmax-A-x3C9-or 4-4-7-xD7-10-x2212-3-2-x3C0-T-or T-9-99-xD7-10-x2212-3-10-xD7-10-x2212-3-0-01s

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7 m m$ , is $4.4 m / s$ . The period of oscillation is
$0.1 s$
$0.01 s$
$100 s$
$10 s$

For a SHM, $x = A sin (ω t + ϕ)$
Velocity, $v = \frac{d x}{d t} = A ω cos (ω t + ϕ)$
So, $v_{m a x} = A ω$
or $4.4 = 7 \times 10^{- 3} (2 π / T)$
or $T = 9.99 \times 10^{- 3} = 10 \times 10^{- 3} = 0.01 s$

The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is 0.1 s0.01 s100 s10 s

For a SHM, x=Asin(ωt+ϕ)Velocity, v=dxdt=Aωcos(ωt+ϕ)So, vmax=Aωor 4.4=7×10−3(2π/T)or T=9.99×10−3=10×10−3=0.01s

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7 m m$ , is $4.4 m / s$ . The period of oscillation is
$0.1 s$
$0.01 s$
$100 s$
$10 s$

For a SHM, $x = A sin (ω t + ϕ)$
Velocity, $v = \frac{d x}{d t} = A ω cos (ω t + ϕ)$
So, $v_{m a x} = A ω$
or $4.4 = 7 \times 10^{- 3} (2 π / T)$
or $T = 9.99 \times 10^{- 3} = 10 \times 10^{- 3} = 0.01 s$