A 0.440-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.440-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.6 cm. (Assume the position of the object is at the origin at t = 0.)

(a) Calculate the maximum value of its speed.
cm/s

(b) Calculate the maximum value of its acceleration.
cm/s2

(c) Calculate the value of its speed when the object is 7.60 cm from the equilibrium position.
cm/s

(d) Calculate the value of its acceleration when the object is 7.60 cm from the equilibrium position.
cm/s2

(e) Calculate the time interval required for the object to move from x = 0 to x = 5.60 cm.
s

Explanation / Answer

part A:

apply KE = EPE

0.5 mv^2 = 0.5 KA^2

where K is spring consatnt

A is amplitude

m is mass

so

speed v^2 =   8 * 0.116*0.116/(0.44)

v^2 = 0.244

v = 0.494 m/s   or 49.4 cm/s

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

appl force F = ma = Kx

so

a = Kx/m

a = 8 * 0.116/0.44

a = -2.11 m/s^2   or 211 cm/s^2

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

spped v^2 = 8 * 00.076 *0.076/0.44

v = 32.4 cm/s

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

a = 8 * 0.76/0.44

a = -1381.8 cm/s^2

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

accleration a at x = 0.056 m

is a = 8 * 0.056/0.44

a    =-101.8 Cm/s^2

apply time period T = (sqrt(4pi^2X/a)

T   = sqrt(4*3.14*3.14 * 0.056)/(1.018)

T = 1.47 secs

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