Question

A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of Ï3 ft/s. Find the times at which the mass is heading downward at a velocity of 3 ft/s.

Answer 1

To find the times at which the mass is heading downward at a velocity of 3 ft/s, we can use the principles of simple harmonic motion. By substituting the known values into the equation, we can solve for the time (t) when the velocity (v) is equal to -3 ft/s.

Step-by-step explanation:

To find the times at which the mass is heading downward at a velocity of 3 ft/s, we can use the principles of simple harmonic motion. The equation for the velocity of an object in simple harmonic motion is v = Aωsin(ωt + φ), where A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase angle.

In this case, the amplitude (A) is 1 foot, the angular frequency (ω) can be calculated using the equation ω = √(k/m) where k is the spring constant (9 lb/ft) and m is the mass of the object (1 slug), and the phase angle (φ) is 0 since the mass is initially released from the equilibrium position (the spring is neither stretched nor compressed).

By substituting the known values into the equation, we can solve for the time (t) when the velocity (v) is equal to -3 ft/s. So the times at which the mass is heading downward at a velocity of 3 ft/s are approximately t = 0.300 seconds and t = 0.694 seconds.

Answer 2

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Step-by-step explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

`m(d^(2) x)/(dt^(2) ) = -kx \\ m(d^(2) x)/(dt^(2) ) + kx = 0\\(d^(2) x)/(dt^(2) ) + (k)/(m) x = 0`

with m = 1 slug and k = 9 lb/ft, the equation becomes

`(d^(2) x)/(dt^(2) ) + (9)/(1) x = 0\\(d^(2) x)/(dt^(2) ) + 9 x = 0`

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ where k is an integer

3t = ±π/6 + 2kπ - 4π/3

t = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number