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Standard XII

Physics

Time Period Independent of Amplitude

From the disp...

Question

From the displacement time graph of an oscillating particle. Find the maximum velocity of the particle :

2 m s^{- 1}

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π m s^{- 1}

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0.2 π m s^{- 1}

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\frac{π}{2} m s^{- 1}

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Solution

The correct option is D $0.2 π m s^{- 1}$

By looking at the graph we can see that amplitude $A = 0.2 m$ and time period $T = 2 s$
and the maximum velocity is given by
$v_{m a x} = ω A = \frac{2 π}{T} A v_{m a x} = \frac{2 π}{2} (0.2) = 0.2 π m s^{- 1}$

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From the displacement time graph of an oscillating particle. Find the maximum velocity of the particle :

The correct option is D 0.2πms−1 By looking at the graph we can see that amplitude A=0.2m and time period T=2sand the maximum velocity is given byvmax=ωA=2πTAvmax=2π2(0.2)=0.2πms−1

The correct option is D $0.2 π m s^{- 1}$

By looking at the graph we can see that amplitude $A = 0.2 m$ and time period $T = 2 s$
and the maximum velocity is given by
$v_{m a x} = ω A = \frac{2 π}{T} A v_{m a x} = \frac{2 π}{2} (0.2) = 0.2 π m s^{- 1}$