Ncert Exercises & Solutions

A particle executing SHM has a maximum speed of $30$ cm/s and a maximum acceleration of $60$ cm/s². The period of oscillation is:
1. $\pi$ s
2. $\dfrac{\pi }{2}$ s
3. $2\pi$ s
4. $\dfrac{\pi }{4}$ s

(1) Hint: The particle will have the maximum speed at the mean position and the maximum acceleration at the extreme position.

Step 1: Find the maximum velocity.

Let equation of an SHM is represented by, y = a sin

ω t

$ω t$

v = \frac{d y}{d t} = a ω \cos ω t

$v = \frac{d y}{d t} = a ω \cos ω t$

\Rightarrow {(v)}_{m a x} = a ω = a ω = 30 . . . (i)

$\Rightarrow {(v)}_{m a x} = a ω = a ω = 30 . . . (i)$

Step 2: Find the maximum acceleration.

Acceleration, (A)

= \frac{d x^{2}}{d t^{2}} = - a ω^{2} \sin ω t

$= \frac{d x^{2}}{d t^{2}} = - a ω^{2} \sin ω t$

A_{m a x} = ω^{2} a = 60 . . . (i i)

$A_{m a x} = ω^{2} a = 60 . . . (i i)$

Step 3: Find the period of the oscillation.

From Eqs. (i) and (ii), we get,

ω (ω a) = 60 \Rightarrow ω (30) = 60

$ω (ω a) = 60 \Rightarrow ω (30) = 60$

\Rightarrow ω = 2 r a d / s

$\Rightarrow ω = 2 r a d / s$

\Rightarrow \frac{2 π}{T} = 2 r a d / s \Rightarrow T = π s e c

$\Rightarrow \frac{2 π}{T} = 2 r a d / s \Rightarrow T = π s e c$

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