A particle executing simple harmonic motion has an angular frequency of $6.28 s^{- 1}$ and amplitude of $10 c m$ . The time period of the SHM is :

1 s

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2 s

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3 s

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4 s

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Solution

The correct option is A $1 s$
Time period = $\frac{2 π}{ω} = \frac{2 π}{6.28} s = 1 s$

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Q. A particle executing simple harmonic motion has an angular frequency of

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A particle executing simple harmonic motion has angular frequency $6.28 s^{-} 1$ and amplitude $10 c m$ . Find $(x)$ the speed when the displacement is $6 c m$ , from the mean position, $(y)$ the speed at $t = \frac{1}{6} s$ assuming that the motion starts from rest at $t = 0$ .
(i) $25.1 c m s^{- 1}$
(ii) $54.4 c m s^{- 1}$
(iii) $50.2 c m s^{- 1}$
(iv) $27.2 c m s^{- 1}$

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A particle executing simple harmonic motion has an angular frequency of 6.28s−1 and amplitude of 10cm. The time period of the SHM is :

The correct option is A 1sTime period = 2πω=2π6.28s=1s

A particle executing simple harmonic motion has an angular frequency of $6.28 s^{- 1}$ and amplitude of $10 c m$ . The time period of the SHM is :

The correct option is A $1 s$
Time period = $\frac{2 π}{ω} = \frac{2 π}{6.28} s = 1 s$