## Trending News

# A mass m = 5.0 kg is attached to a vertical spring with k = 188 N/m and is set into motion.?

A mass m = 5.0 kg is attached to a vertical spring with k = 188 N/m and is set into motion.

(a) What is the frequency of the oscillation?

Hz

(b) If the amplitude of the oscillation is 3.2 cm, what is the maximum value of the velocity?

m/s

(c) How long does it take the mass to move from y = 1.5 cm to y = 2.5 cm?

s

(d) If the mass is oscillating with a maximum speed of 45 m/s, what is the amplitude?

m

(e) If the spring constant is increased by a factor of two and the maximum kinetic energy of the mass is the same, by what factor does the amplitude change?

% of the original amplitude

### 2 Answers

- PearlsawmeLv 71 decade agoFavorite Answer
a)

ω = √ k/m = √ 188/ 5 = 37.6 rad /s. = 2 π N where N is the frequency

N = 37.6 / 2 π Hz = 5.98 Hz.

--------------------------------------------------------------

b)

Maximum velocity = A ω where A is the amplitude.

Maximum velocity = 0.032*37.6 = 1.2032 m/s

---------------------------------------------------------------

c)

y2 = A sin ωt2 and y1 = A sin ωt1.

sin ωt1 = y1/ A= 1.5 / 3.2 = 0.46875

ωt1 = 0.4879

t1 =0.4879 / 37.6 = 0.013 s.

sin ωt2 = y2/ A= 2.5 / 3.2 = 0.78125

ωt2 = 0.8967

t2 = 0.8967/ 37.6 = 0.024 s

The time difference = 0.024 -0.013 = 0.011 s

-----------------------------------------------------------

d)

The max k.e = 1/2 m [ A ω] ^2 is a constant.

that is Aω is a constant.

A1 ω1 = A2 ω2

ω2/ ω1 = A1/A2

ω2/ ω1 = √ k2/k1= A1/A2

Given k2 = 2k1

√ k2/k1 = √2 = A1/A2

Hence A2 = A1 / √2

The factor is 1/√2 = 70.71%

=====================================

- Login to reply the answers

- JoeLv 41 decade ago
(a) f = k/m = 188/5 = 37.6 Hz

(b) x(t) = Asin(f*t) = 3.2sin(37.6t)

v(t) = d/dt(x(t)) = 129.32*cos(37.6t)

cosine maximum = 1, at t = 0 (assuming dampened harmonic oscillator equation):

v(max) = 1.2932 (m/s)

(c) no idea

(d) use eq. in (b) for damp harmonic osc.

(e) no idea

- Login to reply the answers