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3 years ago

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You are watching an object that is moving in SHM. When the object is displaced 0.560 m to the right of its equilibrium position,

it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left. Part A How much farther from this point will the object move before it stops momentarily and then starts to move back to the left

1 answer:

Ilia_Sergeevich [38]3 years ago

3 0

Answer:

2.95m

Explanation:

The farthest distance the object can move is the radius of the circle of which the Simple harmonic motion is assumed to be a part

But V = w× r; where V is velocity,

w is angular velocity and r is radius.

Also,

a= w2r; where a is linear acceleration

but a = v× r ; by comparing both equations

Hence r = a/v =8.6/2.45 =3.51m

But the horizontal distance of the motion is given by:

X = rcosx ; where x is the angle

X is the distance covered.

We know that the maximum value of cos x is 1 which is 0°

When the object moves in a fashion directly parallel to an horizontal distance, maximum distance would be reached and hence:

X = r=3.51m

Meaning the object needs to travel 3.51-0.56=2.95m further.

Note: the acceleration of the motion is constant whether it is swinging towards the left or right.

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Answer:

Explanation:

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3 years ago

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<h2>Answer:</h2>

<em><u>(a). 16.741 m/s</u></em>

<em><u>(b). 15.75 m/s</u></em>

<h2>Explanation:</h2>

In the question,

Height of the cliff, h = 37 m

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Now,

Let us say the speed of the ball = u m/s

So,

Time taken by the ball to reach at the bottom, t is given by,

$t=\sqrt{\frac{2h}{g}}\\t=\sqrt{\frac{2\times 37}{9.8}}\\t=2.747\,s$

So,

Splash is heard after = 2.92 s

So,

<u>Time taken by sound to travel the shortest distance along the hypotenuse of the triangle</u> thus formed is,

t = 2.92 - 2.747

t = 0.172 s

Now,

Distance traveled by sound is given by,

$Distance=343\times 0.172\\Distance=59.02\,m$

So,

In the triangle using the Pythagoras Theorem,

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$D=\sqrt{59.02^{2}-37^{2}}\\D=45.99\,m$

So,

Speed of throwing of ball is given by,

$Distance=Speed\times Time\\45.99=u\times 2.747\\u=16.741\,m/s$

<em><u>Therefore, the speed of the ball = 16.741 m/s.</u></em>

(b).

If,

Speed of sound = 331 m/s

So,

<u>Distance traveled by sound</u> is,

$Distance=331\times 0.172\\Distance=56.932\,m$

So,

Distance traveled in the horizontal by ball is,

$Distance=\sqrt{56.932^{2}-37^{2}}\\Distance=43.269\,m$

So,

Speed of the ball thrown is given by,

$Speed=\frac{Distance}{Time}\\Speed=\frac{43.269}{2.747}\\Speed=15.75\,m/s$

<em><u>Therefore, the speed of the ball = 15.75 m/s.</u></em>

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3 years ago