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Q.A particle executing S.H.M. has a maximum displacement of 4 cm and its acceleration at a distance of 1 cm from its mean position is 3 cms^-2. What will be its velocity when it is at a distance of 2 cm from its mean position?

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Subjective· 2mImportance★★★★★
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Using acceleration = ω²x to find ω, then velocity = ω√(a²−x²), the velocity at x = 2 cm works out to 6 cm/s.

Given: amplitude a=4 cma = 4\ cm; acceleration at x=1 cmx=1\ cm is 3 cm/s23\ cm/s^2.

Step 1 — find ω\omega using the SHM acceleration relation ∣Acc∣=ω2x|A_{cc}| = \omega^2 x:

3=ω2(1)  ⟹  ω2=3  ⟹  ω=3 rad/s3 = \omega^2 (1) \implies \omega^2 = 3 \implies \omega = \sqrt{3}\ \text{rad/s}

Step 2 — find velocity at x=2 cmx = 2\ cm using the SHM velocity relation:

v=ωa2−x2v = \omega\sqrt{a^2-x^2} …

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