A body of mass 500g suspended from the end of a spiral spring which obeys Hooks law, produced an extension of 10cm. If the mass is pulled down a distance of 5cm and released, calculate
1 the force constant of the spring
2 the frequency of oscillation
3 the period of oscillation
4 the angular speed of the body
1. The force constant (k) of the spring can be calculated using Hooke's Law, which states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position. The formula for Hooke's Law is:
F = k * x
Given that the extension is 10 cm (0.10 m) and the mass is 500 g (0.5 kg), and considering that acceleration due to gravity is approximately 9.8 m/s², we can solve for the force constant:
500 g * 9.8 m/s² = k * 0.10 m
k = (500 g * 9.8 m/s²) / 0.10 m
k = 490 N/m
Therefore, the force constant of the spring is 490 N/m.
2. The frequency of oscillation (f) can be calculated using the formula:
f = (1 / 2π) * √(k / m)
Given that k = 490 N/m (force constant) and m = 0.5 kg (mass), we can calculate the frequency:
f = (1 / 2π) * √(490 N/m / 0.5 kg)
f = (1 / 2π) * √(980 N/kg)
f ≈ (1 / 2π) * 31.30 s⁻¹
f ≈ 4.98 Hz
Therefore, the frequency of oscillation is approximately 4.98 Hz.
3. The period of oscillation (T) is the time taken for one complete oscillation and can be calculated using the formula:
T = 1 / f
Given that f = 4.98 Hz (frequency), we can calculate the period:
T = 1 / 4.98 s⁻¹
T ≈ 0.20 seconds
Therefore, the period of oscillation is approximately 0.20 seconds.
4. The angular speed of the body (ω) can be calculated using the formula:
ω = 2π * f
Given that f = 4.98 Hz (frequency), we can calculate the angular speed:
ω = 2π * 4.98 s⁻¹
ω ≈ 31.30 rad/s
Therefore, the angular speed of the body is approximately 31.30 rad/s.
To calculate the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
1. Force constant (k):
According to Hooke's Law, F = kx, where F is the force applied, k is the force constant, and x is the displacement. In this case, the body of mass 500g (0.5 kg) produces an extension of 10 cm (0.1 m). The weight of the body can be calculated as follows:
Weight (W) = mass × acceleration due to gravity
W = 0.5 kg × 9.8 m/s^2
W = 4.9 N
Since the weight is balanced by the force exerted by the spring, we can say that F = W. Therefore, we have:
k × 0.1 m = 4.9 N
k = 4.9 N / 0.1 m
k = 49 N/m
So, the force constant of the spring is 49 N/m.
2. Frequency of oscillation:
The frequency (f) of oscillation of a mass-spring system is given by:
f = (1 / (2π)) * √(k / m)
where π is a mathematical constant, k is the force constant of the spring, and m is the mass of the body.
Using the values k = 49 N/m and m = 0.5 kg, we can calculate the frequency:
f = (1 / (2π)) * √(49 N/m / 0.5 kg)
f = (1 / (2π)) * √(98 N/kg)
f ≈ 3.13 Hz (rounded to two decimal places)
So, the frequency of oscillation is approximately 3.13 Hz.
3. Period of oscillation:
The period (T) of oscillation is the reciprocal of the frequency and can be calculated using the following formula:
T = 1 / f
Substituting the value of f calculated above, we get:
T = 1 / 3.13 Hz
T ≈ 0.32 s (rounded to two decimal places)
So, the period of oscillation is approximately 0.32 seconds.
4. Angular speed (ω):
The angular speed (ω) of the body is commonly represented in radians per second (rad/s) and can be calculated using the following formula:
ω = 2πf
Substituting the value of f calculated earlier, we get:
ω = 2π * 3.13 Hz
ω ≈ 19.68 rad/s (rounded to two decimal places)
So, the angular speed of the body is approximately 19.68 rad/s.
To solve this problem, we need to use the concepts of Hooke's Law and simple harmonic motion.
1. The force constant of the spring (k) can be determined using Hooke's Law formula: F = -kx, where F is the force applied, k is the force constant, and x is the extension or displacement of the spring. In this case, the force applied is the weight of the body (mg), where m is the mass and g is the acceleration due to gravity. The extension of the spring is given as 10 cm, which is equivalent to 0.1 m. Thus, we can write the equation as:
mg = kx
0.5 kg * 9.8 m/s^2 = k * 0.1 m
Simplifying the equation, we can find the force constant:
k = (0.5 kg * 9.8 m/s^2) / 0.1 m
k = 49 N/m
Therefore, the force constant of the spring is 49 N/m.
2. The frequency of oscillation (f) can be calculated using the formula: f = 1 / T, where T is the period of oscillation. The period (T) is the time taken for one complete cycle of oscillation. To find the period, we can use the formula for the period of simple harmonic motion: T = 2π√(m/k), where m is the mass and k is the force constant.
T = 2π√(0.5 kg / 49 N/m)
Calculating T, we get:
T ≈ 2.54 s
Substituting the period into the frequency formula, we have:
f = 1 / T
f = 1 / 2.54 s
f ≈ 0.394 Hz
Therefore, the frequency of oscillation is approximately 0.394 Hz.
3. The period of oscillation (T) has already been calculated in the previous step. It is approximately 2.54 seconds.
4. The angular speed of the body (ω) can be calculated using the formula: ω = 2πf, where f is the frequency of oscillation. Since we calculated the frequency as 0.394 Hz, we can substitute it into the formula:
ω = 2π * 0.394 rad/s
ω ≈ 2.47 rad/s
Therefore, the angular speed of the body is approximately 2.47 rad/s.
m = 500 g = 0.500 kg
x = 10 cm = 0.10 m
g = 9.81 m/s^2
F = k x
0.5 * 9.81 = k * 0.10
k = 49 Newtons / meter
F = m a
-k x = m a
if x = A sin (2 pi f t)
v = A (2 pi f) cos (2 pi f t)
a = -A (2 pi f)^2 sin (2 pi f t) = - (2 pi f)^2 x
so
-k = -(2 pi f)^2
(2 pi f)^2 =49
2 pi f = 7 = omega by the way
f = 7 / 2pi = 1.11 Hz
T =1/f
omega = 2 pi f = 7 radians/ second