A child bounces in a harness suspended from a door frame by three elastic bands.
(a) If each elastic band stretches 0.210 m while supporting a 7.35-kg child, what is the force constant for each elastic band?
114.33 N/m
(b) What is the time for one complete bounce of this child?
0.9198s
(c) What is the child's maximum velocity if the amplitude of her bounce is 0.210 m?
Oh, I'd say the child's maximum velocity is going to be quite bouncy! Maybe they'll turn into a little Superman or Supergirl, flying through the air at superspeed! Hang tight, kiddo!
To find the force constant, you can use Hooke's law, which states that the force exerted by an elastic object is proportional to the displacement from its equilibrium position.
(a) Force constant for each elastic band:
1. Calculate the force applied on each elastic band:
F = mass * gravitational acceleration
F = 7.35 kg * 9.8 m/s^2
F = 71.913 N
2. Use Hooke's law to find the force constant:
F = k * x
where F is the force applied, k is the force constant, and x is the displacement.
Rearranging the equation, we have:
k = F / x
k = 71.913 N / 0.210 m
k ≈ 114.33 N/m
Therefore, the force constant for each elastic band is 114.33 N/m.
(b) Time for one complete bounce:
The time for one complete bounce can be calculated using the following formula:
T = 2π * √(m / k)
where T is the period, m is the mass, and k is the force constant.
Substituting the given values into the formula, we have:
T = 2π * √(7.35 kg / 114.33 N/m)
T ≈ 2π * √(0.064 kg•m/N)
T ≈ 2π * 0.2534 s
T ≈ 1.591 s
Therefore, the time for one complete bounce is approximately 1.591 seconds.
(c) Maximum velocity of the child:
The maximum velocity of the child can be calculated using the equation for simple harmonic motion:
v_max = ω * A
where v_max is the maximum velocity, ω is the angular frequency, and A is the amplitude.
The angular frequency (ω) can be found using the formula:
ω = √(k / m)
where k is the force constant and m is the mass.
Substituting the known values into the formula, we have:
ω = √(114.33 N/m / 7.35 kg)
ω ≈ √15.57 s^(-2)
ω ≈ 3.95 s^(-1)
Finally, substituting ω and A into the equation for maximum velocity:
v_max = 3.95 s^(-1) * 0.210 m
v_max ≈ 0.828 m/s
Therefore, the child's maximum velocity is approximately 0.828 m/s.
To answer these questions, we need to use Hooke's Law and the formula for the period of a simple harmonic motion.
Hooke's Law states that the force exerted by an elastic material is directly proportional to the displacement from its equilibrium position. Mathematically, this relationship is expressed as:
F = k * x
where F is the force, k is the force constant (also called the spring constant), and x is the displacement from the equilibrium position.
(a) To find the force constant for each elastic band, we need to calculate it using the given information. We know that each elastic band stretches 0.210 m while supporting a 7.35-kg child. So, we have:
F = k * x
Rearranging the formula, we get:
k = F / x
Substituting the values, we have:
k = (m * g) / x
where m is the mass of the child (7.35 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²). Plugging in the values, we get:
k = (7.35 kg * 9.8 m/s²) / 0.210 m
k ≈ 114.33 N/m
Therefore, the force constant for each elastic band is approximately 114.33 N/m.
(b) The period of a simple harmonic motion can be calculated using the formula:
T = 2π * √(m/k)
where T is the period, m is the mass, and k is the force constant.
Substituting the values, we have:
T = 2π * √(m/k)
T = 2π * √(7.35 kg / 114.33 N/m)
Simplifying this equation, we get:
T ≈ 2π * √(0.064 kg * m/N)
Evaluating the expression, we find:
T ≈ 2π * 0.253 s
T ≈ 1.59 s
Therefore, the time for one complete bounce of the child is approximately 1.59 seconds.
(c) The maximum velocity (v) of the child can be determined using the formula:
v = A * ω
where A is the amplitude (0.210 m) and ω is the angular frequency, given by:
ω = √(k/m)
Substituting the values, we have:
ω = √(k / m)
ω = √(114.33 N/m / 7.35 kg)
Simplifying this equation, we get:
ω ≈ √(15.55 N/m * kg)
Evaluating the expression, we find:
ω ≈ 3.94 rad/s
Now, we can calculate the maximum velocity:
v = A * ω
v = 0.210 m * 3.94 rad/s
Evaluating the expression, we find:
v ≈ 0.826 m/s
Therefore, the child's maximum velocity is approximately 0.826 m/s.
F = k *stretch for each
each supports 7.35 * 9.81/3
= 24 N
24 = k (.21)
k = 115 N/m agree except you use 9.8 for g I suspect
K = 3 k from part a = 343
omega = 2 pi f = sqrt(K/m)
= 6.83 radians/s
f = omega/2pi = 1.09 Hz
T = 1/f = .919 seconds agree
V = A omega sin (omega t)
max of sin is 1
Vmax = A omega = .210*6.83
= 1.44 m/s