The following figure shows the displacement versus time graph for two particles $A$ and $B$ executing simple harmonic motions. The ratio of their maximum velocities (respectively) is

3 : 1

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1 : 3

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1 : 9

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9 : 1

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Solution

The correct option is A $3 : 1$
For $A$ , time period $T_{A} = 16 s$ (distance between two adjacent crests)
For $B$ , time period $T_{B} = 2 (20 - 8) = 24 s$
(length between the crest and trough shown $= 20 s - 8 s = 12 s$ )
Also, amplitudes $a_{A} = 10 cm, a_{B} = 5 cm$
Now, $\frac{(V_{max})_{A}}{(V_{max})_{B}} = \frac{ω_{A} a_{A}}{ω_{B} a_{B}}$
$\frac{(\frac{2 π}{T_{A}}) a_{A}}{(\frac{2 π}{T_{B}}) a_{B}} = \frac{T_{B} a_{A}}{T_{A} a_{B}} = \frac{24 \times 10}{16 \times 5} = \frac{3}{1}$

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The following figure shows the displacement versus time graph for two particles A and B executing simple harmonic motions. The ratio of their maximum velocities (respectively) is

The correct option is A 3:1For A, time period TA=16 s (distance between two adjacent crests) For B, time period TB=2(20−8)=24 s (length between the crest and trough shown=20 s−8 s=12 s) Also, amplitudes aA=10 cm,aB=5 cm Now, (Vmax)A(Vmax)B=ωAaAωBaB (2πTA)aA(2πTB)aB=TBaATAaB=24×1016×5=31

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