A particle in SHM has an amplitude of 20 cm and time period of 2 sec. Its maximum velocity will be(in m/s):

0.04 π^{2}

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0.2 π

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0.2 π^{2}

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0.04 π

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Solution

The correct option is B $0.2 π$

$v = \frac{d x}{d t} = A ω c o s (ω t + θ)$
Maximum $v$ occurs for maximum value of $c o s (ω t + θ)$ , which is 1
Thus, maximum $v = A ω$
Now, $ω = \frac{2 π}{T} = \frac{2 π}{2} = π$ , and $A = 20 c m = 0.2 m$
Thus, maximum v = $0.2 π$

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A particle in SHM has an amplitude of 20 cm and time period of 2 sec. Its maximum velocity will be(in m/s):

The correct option is B 0.2 π v=dxdt=Aωcos(ωt+θ)Maximum v occurs for maximum value of cos(ωt+θ), which is 1Thus, maximum v=AωNow, ω=2πT=2π2=π, and A=20cm=0.2mThus, maximum v = 0.2π

The correct option is B $0.2 π$

$v = \frac{d x}{d t} = A ω c o s (ω t + θ)$
Maximum $v$ occurs for maximum value of $c o s (ω t + θ)$ , which is 1
Thus, maximum $v = A ω$
Now, $ω = \frac{2 π}{T} = \frac{2 π}{2} = π$ , and $A = 20 c m = 0.2 m$
Thus, maximum v = $0.2 π$