Question

A block with mass m =7.1 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.24 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.5 m/s. The block oscillates on the spring without friction.

1) What is the spring constant of the spring?

2) What is the oscillation frequency?

3) After t = 0.38 s what is the speed of the block?

4) What is the magnitude of the maximum acceleration of the block?

Answer #1

**given
m = 7.1 kg
x = 0.24 m
v_max = 4.5 m/s**

**1) In the equilibrium, F_spring = m*g**

**k*x = m*g**

**k = m*g/x**

**= 7.1*9.8/0.24**

**= 290 N/m
<<<<<<<<<<---------------Answer**

**2) we know, angular frequency of motion, w =
sqrt(k/m)**

**= sqrt(290/7.1)**

**= 6.391 rad/s**

**we know, f = w/(2*pi)**

**= 6.391/(2*pi)**

**= 1.02 Hz
<<<<<<<<<<---------------Answer**

**3) v = vmax*cos(w*t)**

**= 4.5*cos(6.391*0.38) (use calculator in radian
mode)**

**= 3.40 m/s
<<<<<<<<<<---------------Answer**

**4) let A is the amplitude of motion.**

**v_max = A*w**

**and**

**a_max = A*w^2**

**= (A*w)*w**

**= v_max*w**

**= 4.5*6.391**

**= 28.8 m/s^2
<<<<<<<<<<---------------Answer**

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