A 4.5 kg mass is connected to a spring (k-185 N/m) and is sliding on a horizonta

Question

A 4.5 kg mass is connected to a spring (k-185 N/m) and is sliding on a horizontal frictionless surface. The mass is given an initial displacement of +20 cm and released with an initial velocity of -19 cm/s. Determine the acceleration of the spring at t 2.0 seconds. (include units with answer) -13.10 m/sec" 4 pts,96% # tries: 2 show Details Format Check 296 try penalty Hints: 1,1 A simple pendulum has a period of 1.14 sec. If the pendulum's length is decreased by 13.1 cm, what is the new period? (include units with answer) 4.17 pts 100% B. # tries: 0 Show Details Format Check 2% try penalty Hints: 0,0 t W 82 cm H H 49 cm X 13 cm VOLS 4.17 pts. 100% # tries: 0 show Details C. One connection breaks on this rectangular sign and it falls and swings about point A. How long does it take to complete one oscillation? (include units with answer) Format Check 296 try penalty Hints: 11 L 1.8 m x = 10.1 cm y-3.9 cm M55 kg 4.17 pts 100% # tries: 0 Show Details A 2.2 mm diameter, 1.8 m long cord Format Check (p-7800 kg/m3) is tensioned between a wall support and a pivot arm supporting a 55 kg mass. What is the cord's fundamental frequency of vibration for a transverse wave? 2% try penalty Hints: 0,2

Explanation / Answer

(A) w = sqrt( k /m ) = sqrt(185 / 4.5) = 6.41 rad/s

x = A cos(wt + phi)

at t = 0

20 = A cos(phi) ..... (i)

and 19 = A w sin(phi) ..... (ii)

(ii)/(i) => w tan(phi) = 19/20

phi = 0.147 rad

so A = 20.22 cm

a = - A w^2 cos(wt + phi)

a = - (0.2022 x 6.41^2) cos((6.421 x 2) + 0.147)

a = - 7.64 m/s^2 .........Ans

B. T = 2 pi sqrt[ L / g]

1.14 = 2 pi sqrt(L / 9.8)

L = 0.3226 m

L' = 0.3226- 0.131 = 0.1916 m

T' = 2 pi sqrt[ 0.1916 / 9.8] = 0.878 sec ...........Ans

C. I = m(H^2 + w^2) / 12 = m (0.49^2 + 0.82^2)/12 = 0.076 m

and m g d = m g sqrt[(0.49/2)^2 + (0.41 - 0.13)^2 ] = 0.372 m g

T = 2 pi sqrt[ 0.076 / (0.372 x 9.8)] = 0.907 sec

D. (x M g) = (T y)

T = (10.1 x 55 x 9.8) / 3.9 = 1396 N

linear mass density = (7800)(pi (1.1 x 10^-3)^2) = 0.02965 kg/m

v = sqrt[ T / linear mass density] = 217 m/s

wavelength = 2L = 3.6 m

f = v / wavelength = 60.3 Hz

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