Question

Q.5.* A particle executes S.H.M. along a straight line with mean position x = 0, period 20 s and amplitude 5 cm. The shortest time taken by the particle to go from x = 4 cm to x = - 3 cm is
(A) 4 s
(B) 7 s
(C) 5 s
(D) 6 s

Answer 1

Answer:

This can be solved by using phaser diagram at x = 4cm angle is 53 and at -3 cm angle is -37 shortest angle between them is 90 . for 360 degree it takes 20 seconds so to travel 90 degree it will take 20/4= 5 seconds

Answer 2

The shortest time is 5 sec.

(C) is correct option.

Explanation:

Given that,

Period = 20 s

Amplitude = 5 cm

The particle to go from x = 4 cm to x = - 3 cm

So the minimum phase difference between two position

$\theta=\theta_{1}+\theta_{2}$

Put the value into the formula

$\theta=53+37$

$\theta=90^{\circ}$

We need to calculate the time

Using formula of time

$\theta=\omega t$

$t=\dfrac{\theta}{\omega}$

Put the value into the formula

$t=\dfrac{\dfrac{\pi}{2}}{\dfrac{2\pi}{20}}$

$t=5\ sec$

Hence, The shortest time is 5 sec.

Learn more :

Topic : simple harmonic motion

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