A 190 g mass attached to a horizontal spring oscillates at a frequency of 5.50 H

Question

A 190 g mass attached to a horizontal spring oscillates at a frequency of 5.50 Hz . At t =0s, the mass is at x= 6.80 cm and has vx = 32.0 cm/s . Determine:

Part A

The period.

Express your answer to three significant figures and include the appropriate units.

0.182 s

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Correct

Part B

The angular frequency.

Express your answer to three significant figures and include the appropriate units.

34.6 rads

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Correct

Part C

The amplitude.

Express your answer to three significant figures and include the appropriate units.

6.86×102 m

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Correct

Part D

The phase constant.

Express your answer to three significant figures and include the appropriate units.

0.135 rad

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Correct

Part E

The maximum speed.

Express your answer to three significant figures and include the appropriate units.

2.37 ms

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Correct

Part F

The maximum acceleration.

Express your answer to three significant figures and include the appropriate units.

82.0 ms2

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Correct

Part G

The total energy.

Express your answer to three significant figures and include the appropriate units.

0.534 J

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Part H

The position at t= 3.20 s .

Express your answer to three significant figures and include the appropriate units.

* I DON'T KNOW HOW TO FIND PART H!

0.182 s

Explanation / Answer

(a) Period is the inverse of the frequency.

Period = 1 / (5.50 Hz )
= 0.181seconds ---



(b) = 2f
= 2( 5.50 Hz )
= 34.55 rad / sec ---



(c) Max potential = kinetic + potential
½kA² = ½kx² + ½mv²
kA² = kx² + mv²
A² = x² + mv² / k
A² = x² + mv² / m²
A² = x² + ( v/ )²
A² = ( 6.80 cm )² + ( -32 cm/s /34.55 s¹ )²
A² =45.31 cm²
A =6.73cm ---



(d) velocity = v(t)
v(t) = A·cos( t + )
-32 cm/s = ( 6.73 cm )( 34.55 s^-1 )cos( 0 + )
-7.26 = cos( )
= cos¹( -.7.26 )
=.135 radians ---



(e) Max speed = A
= ( 6.73 cm )( 34.55 s¹ )
=232.52 cm/s

=2.32 m/s---

(f) Max acceleration = A²
= ( 6.73 cm )( 34.55 s¹ )²
= 8033.61 cm/s²

= 82.0---



(g) Total energy = max kinetic energy
= ½m( max speed )²
= ½ ( 0.19 kg )( 2.32 m/s )²
= 4408 kg cm²/s²
= 0.04408 joules ---



(h) Position = s(t)
= A·sin[ t + ]
= ( 6.73 cm )sin[ (34.55 s¹)( 3.20 s ) +.135 ]
=
=6.29cm ---

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