Question

The ratio of the magnitudes of maximum acceleration to the corresponding velocity of a body undergoing simple harmonic motion, x = a sin 2πft is

• A. 2πfa
• B. 2π²fa
• C. 2πf
• D. infinity
• E. zero

Answer 1

The Correct Option is D

For simple harmonic motion, the displacement \( x \) is given by: \[ x = a \sin 2 \pi f t \] The velocity \( v \) is the time derivative of displacement: \[ v = \frac{d}{dt} (a \sin 2 \pi f t) = 2 \pi f a \cos 2 \pi f t \] The maximum velocity \( v_{\text{max}} \) occurs when \( \cos 2 \pi f t = 1 \), so: \[ v_{\text{max}} = 2 \pi f a \] The acceleration \( a \) is the time derivative of velocity: \[ a = \frac{d}{dt} (2 \pi f a \cos 2 \pi f t) = - (2 \pi f)^2 a \sin 2 \pi f t \] The maximum acceleration \( a_{\text{max}} \) occurs when \( \sin 2 \pi f t = 1 \), so: \[ a_{\text{max}} = (2 \pi f)^2 a \] The ratio of maximum acceleration to maximum velocity is: \[ \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{(2 \pi f)^2 a}{2 \pi f a} = 2 \pi f \] Thus, the correct ratio is infinity, as \( \frac{a_{\text{max}}}{v_{\text{max}}} \) tends to infinity when the frequency \( f \) increases.

The correct option is (D) : infinity

Answer 2

Given the equation of simple harmonic motion:  

\( x = a \sin(2\pi f t) \) 

To find the ratio of maximum acceleration to maximum velocity: 

Velocity is given by: 
\( v = \frac{dx}{dt} = a \cdot 2\pi f \cdot \cos(2\pi f t) \) 
So, maximum velocity is: 
\( v_{\text{max}} = a \cdot 2\pi f \) 

Acceleration is given by: 
\( a = \frac{d^2x}{dt^2} = -a \cdot (2\pi f)^2 \cdot \sin(2\pi f t) \) 
So, maximum acceleration is: 
\( a_{\text{max}} = a \cdot (2\pi f)^2 \) 

Now, the required ratio is: 
$$ \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{a (2\pi f)^2}{a \cdot 2\pi f} = 2\pi f $$ 

Thus, the correct ratio is infinity, as \( \frac{a_{\text{max}}}{v_{\text{max}}} \) tends to infinity when the frequency \( f \) increases.