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Relating Circular Motion to SHM

The maximum v...

Question

The maximum velocity of simple harmonic motion represented by $y = 3 sin (100 t + \frac{π}{6})$ is given by

300

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\frac{3 π}{6}

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100

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\frac{π}{6}

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Solution

The correct option is B $300$
From the equation of the wave,
Amplitude of wave is $A = 3$
The angular velocity of the wave can be obtained as $ω = 100 r a d / s e c$

We know that the velocity can be obtained as:
$v_{m a x} = a ω$

$= 3 \times 100 = 300$

Suggest Corrections

Similar questions

x_{1} = A sin (ω t + \frac{π}{6})

x_{2} = A cos (ω t)

y_{1} = 0.1 sin (100 π t + \frac{π}{3})

y_{2} = 0.1 cos π t

1

2

2 kg

y (in cm)

t

y = 6 sin (200 t + \frac{π}{6})

2 kg

y (in cm)

t

y = 6 sin (200 t + \frac{π}{6})

y_{1} = 0.1 s i n (100 π t + \frac{π}{3})

y_{2} = 0.1 c o s π t

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