Ncert Exercises & Solutions

What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?

Hint: The acceleration is maximum at the extreme position and the velocity is maximum at the mean position.

Step 1: Find the maximum velocity.

Let x = Asin

ω t

is the displacement function of SHM.

Velocity,

v = \frac{d x}{d t} = A ω \cos ω t

v_{m a x} = A ω | \cos ω t |_{m a x}

= A ω \times 1 = ω A [∵ | \cos ω t |_{m a x} = 1] . . . (i)

Acceleration,

a = \frac{d v}{d t} = - ω A \cdot ω \sin ω t

= - ω^{2} A \sin ω t

| a_{m a x} | = | (- ω^{2} A) (+ 1) | [∵ {(\sin ω t)}_{m a x} = 1]

| a_{m a x} | = ω^{2} A . . . (i i)

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

\frac{v_{m a x}}{a_{m a x}} = \frac{ω A}{ω^{2} A} = \frac{1}{ω}

\Rightarrow \frac{a_{m a x}}{v_{m a x}} = ω