The figure shows a graph between velocity and displacement (from mean position) of a particle performing SHM. select the incorrect statement

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the time period of the particle is 2s

the maximum acceleration will be $40 c m / s^{2}$

the velocity of particle is $2 \sqrt{21}$ cm/s when it is at a distance 1 cm from the mean position.

Amplitude of SHM is 2.5 cm.

answer is A.

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Detailed Solution

$V = ω A a n d ω = \frac{10}{2.5} = 4$

$(1) T = \frac{2 π}{ω} = \frac{π}{2} = 1.57$

$(2) a = ω^{2} A = 40$

$(3) V = ω \sqrt{A^{2} - X^{2}} = 2 \sqrt{21}$

$(4) A = 2.5 c m$

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