1. A 1.45 kg mass oscillates according to the equation x =0.700cos8.40 t , where

Question

1.

A 1.45 kg mass oscillates according to the equation x=0.700cos8.40t, where x is in meters and t is in seconds.

PART A: Determine the amplitude.

PART B: Determine the frequency.

PART C: Determine the total energy.

PART D: Determine the kinetic energy when x = 0.400 m .

PART E: Determine the potential energy when x = 0.400 m

2.

An object with mass 3.6 kg is executing simple harmonic motion, attached to a spring with spring constant 310 N/m . When the object is 0.024 m from its equilibrium position, it is moving with a speed of 0.40 m/s .

PART A: Calculate the amplitude of the motion.
PART B: Calculate the maximum speed attained by the object.

Explanation / Answer

1. The equation of motion is:
x=0.700cos8.40t

A) Amplitude, A = 0.700 m

B) Frequency, N = w/ 2 = 8.4/ 2 = 1.34 Hz

C) Total Energy, TE = KE + PE = 25.07 J + 16.88 J = 41.95 J (using KE & PE from below options C and D)

D) Kinetic Energy at x = 0.4 m,

KE = 1/2 m ^2 A^2 = 1/2*1.45*(8.4^2)*(0.7^2) = 25.07 J

E) Potential Energy at x = 0.4 m,

PE = 1/2 m ^2 (A^2 - x^2) = 1/2*1.45*(8.4^2)*(0.7^2 - 0.4^2) = 16.88 J

2.

A) Velocity = (A^2-x^2)

Angular frequency = (k/m)
= (310/3.6) = 86.11

Velocity v = 0.4 = 86.11(A^2-(0.024)^2)

Squaring both sides we get, 0.16 = 86.11(A^2-(0.000576)
=> A^2 = 24.34 x 10^-4
=> Amplitude A = 24.34 x 10^-2 m = 4.9 x 10^-2 m

B) Maximum velocity = A = 86.11 x 4.9 x 10^-2 = 0.45 m/s

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