If a 60 kg person stands on a low wall with her full weight on the ball of one foot and the heel free to move, the stretch of the Achilles tendon will cause her center of gravity to lower by about 2.5 mm. What is the spring constant of her Achilles tendon? If she bounces a little, what is her oscillation period?

F=-kx

mg=-k(0.0025)
60*-9.8=k(0.0025
k=2.4*10^5

To find the spring constant of the Achilles tendon, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.

According to the information given, the stretch of the Achilles tendon causes the person's center of gravity to lower by about 2.5 mm. We can consider this displacement as the elongation of the spring.

First, let's convert the displacement from millimeters to meters: 2.5 mm = 2.5/1000 = 0.0025 m.

Next, we need to determine the force exerted by the Achilles tendon. The weight of the person is acting downward, and the elongation of the Achilles tendon is opposing this force. Therefore, the force exerted by the Achilles tendon is equal to the weight of the person.

The weight can be calculated using the formula: Weight = mass x gravity, where the mass is given as 60 kg and the acceleration due to gravity is approximately 9.8 m/s^2.

Weight = 60 kg x 9.8 m/s^2 = 588 N.

Since the force exerted by the Achilles tendon is equal to the weight, we can say that k (the spring constant) multiplied by the elongation (0.0025 m) is equal to 588 N.

k * 0.0025 m = 588 N.

Solving for k: k = 588 N / 0.0025 m = 235,200 N/m.

Therefore, the spring constant of the Achilles tendon is approximately 235,200 N/m.

Now, let's calculate the oscillation period if the person bounces a little. The oscillation period can be found using the formula: T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.

Given that the mass of the person is 60 kg and the spring constant of the Achilles tendon is 235,200 N/m, we can substitute these values into the formula.

T = 2π√(60 kg / 235,200 N/m).

Calculating this equation gives us: T ≈ 2π√(0.000255102).

T ≈ 2π * 0.015979.

T ≈ 0.1005336 seconds.

Therefore, the oscillation period is approximately 0.101 seconds.

To find the spring constant of the Achilles tendon, we can use Hooke's Law which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. The formula for Hooke's Law is:

F = -k*x

Where:
F is the force exerted by the spring,
k is the spring constant,
x is the displacement from the equilibrium position.

In this case, the center of gravity lowers by 2.5 mm due to the stretch of the Achilles tendon. We can assume this displacement as x. The weight of the person is acting as the force exerted by the spring. Therefore, we have:

F = mg

Where:
m is the mass of the person (60 kg),
g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the values, we have:

mg = -k*x

Rearranging the equation, we get:

k = -mg/x

Now, plugging in the values, we get:

k = -(60 kg)(9.8 m/s^2)/(2.5 mm)

To calculate k, we need to convert the displacement from millimeters to meters, as the SI unit for the spring constant is N/m. 2.5 mm is equal to 0.0025 meters.

k = -(60 kg)(9.8 m/s^2)/(0.0025 m)

Now, we can calculate k:

k ≈ -23520 N/m

So, the spring constant of the person's Achilles tendon is approximately 23520 N/m.

Now, let's move on to calculating the oscillation period if the person bounces a little. The oscillation period of a mass-spring system can be determined using the formula:

T = 2π√(m/k)

Where:
T is the period of oscillation,
m is the mass of the person (60 kg),
k is the spring constant (23520 N/m).

Plugging in the values, we get:

T = 2π√(60 kg / 23520 N/m)

Now, we can calculate T:

T ≈ 2π√(0.00255 s)

T ≈ 0.321 s

Therefore, the oscillation period is approximately 0.321 seconds.