A coiled Hookean spring is stretched 10 cm when a 1.5kg mass is hung from it. Suppose a 4 kg mass hangs from the spring and is set into vibration with an amplitude of 12 cm.
A. Find the force constant of the spring.
B. Find the Maximum restoring force acting on the vibrating body.
C. Find the period of vibration
D. Find maximum velocity and maximum acceleration of the vibrating object
E. Find the speed and acceleration when the displacement is 9cm
a)kx^2=mg k=mg/x=1.5*9.8/0.1=147N/m
b)f=kx=147*.12=17.64N
c)T=2pi*sqrt(m/k)=2pi*sqrt(4/147)=1
d)Vmax=Aw w=sqrt(k/m)=6.1 Vmax=6.1*.12=.73m/s
amax=-Aw^2=-.12*6.1^2=-4.4m/s^2
e)V=L/t
A. To find 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. The equation for Hooke's Law is:
F = -kx
where F is the force, k is the force constant, and x is the displacement.
Given that the spring is stretched 10 cm (0.10 m) when a 1.5 kg mass is hung from it, we can substitute these values into the equation to solve for k:
1.5 kg * 9.8 m/s^2 = k * 0.10 m
Simplifying the equation:
k = (1.5 kg * 9.8 m/s^2) / 0.10 m
k = 147 N/m
Therefore, the force constant of the spring is 147 N/m.
B. The maximum restoring force acting on the vibrating body can be found by multiplying the force constant by the amplitude of vibration. The equation is:
F_max = k * A
Given that the amplitude of vibration is 12 cm (0.12 m), we can substitute these values into the equation:
F_max = 147 N/m * 0.12 m
F_max = 17.64 N
Therefore, the maximum restoring force acting on the vibrating body is 17.64 N.
C. The period of vibration can be calculated using the formula:
T = 2π * √(m/k)
where T is the period, m is the mass of the vibrating object, and k is the force constant of the spring.
Given that the mass of the vibrating object is 4 kg and the force constant of the spring is 147 N/m, we can substitute these values into the equation:
T = 2π * √(4 kg / 147 N/m)
T = 2π * √(0.027 m/kg)
T = 2π * 0.164 s
T ≈ 1.031 s
Therefore, the period of vibration is approximately 1.031 seconds.
D. The maximum velocity and maximum acceleration of the vibrating object can be determined using the equations:
v_max = A * ω
a_max = A * ω^2
where v_max is the maximum velocity, a_max is the maximum acceleration, A is the amplitude of vibration, and ω is the angular frequency.
The angular frequency can be calculated using the equation:
ω = √(k / m)
Given that the force constant of the spring is 147 N/m and the mass of the vibrating object is 4 kg, we can substitute these values into the equation:
ω = √(147 N/m / 4 kg)
ω = √(36.75 N/m)
ω ≈ 6.062 rad/s
Now, we can use the angular frequency to calculate the maximum velocity and maximum acceleration:
v_max = 0.12 m * 6.062 rad/s
v_max ≈ 0.727 m/s
a_max = 0.12 m * (6.062 rad/s)^2
a_max ≈ 2.185 m/s^2
Therefore, the maximum velocity of the vibrating object is approximately 0.727 m/s, and the maximum acceleration is approximately 2.185 m/s^2.
E. To find the speed and acceleration when the displacement is 9 cm (0.09 m), we can use the equations for velocity and acceleration:
v = A * ω * cos(ωt)
a = -A * ω^2 * sin(ωt)
Given that the amplitude of vibration is 12 cm (0.12 m) and the displacement is 9 cm (0.09 m), we can substitute these values into the equations for velocity and acceleration:
v = 0.12 m * 6.062 rad/s * cos(6.062 rad/s * t)
a = -0.12 m * (6.062 rad/s)^2 * sin(6.062 rad/s * t)
To find the speed and acceleration at the displacement of 0.09 m, we need to find the value of t that corresponds to that displacement. For simple harmonic motion, the equation for displacement as a function of time is:
x = A * cos(ωt)
Given that the amplitude of vibration is 12 cm (0.12 m) and the displacement is 9 cm (0.09 m), we can substitute these values into the equation and solve for t:
0.09 m = 0.12 m * cos(6.062 rad/s * t)
cos(6.062 rad/s * t) = 0.09 m / 0.12 m
cos(6.062 rad/s * t) = 0.75
Using the inverse cosine function, we can find the value of t:
6.062 rad/s * t = cos^(-1)(0.75)
t = cos^(-1)(0.75) / 6.062 rad/s
Solving this equation gives us the value of t. Plugging this value into the equations for velocity and acceleration will give us the speed and acceleration when the displacement is 9 cm (0.09 m).
To find the answers to these questions, we need to use the concept of Hooke's Law and Simple Harmonic Motion.
A. Force constant of the spring:
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it can be expressed as F = -kx, where F is the force, k is the force constant, and x is the displacement.
In this case, when a 1.5kg mass hangs from the spring and stretches it by 10 cm (0.1 m), the force exerted by the spring can be calculated using the equation F = mg, where m is the mass and g is the acceleration due to gravity (approximated as 9.8 m/s^2). So, F = (1.5 kg)(9.8 m/s^2) = 14.7 N.
Now, plug in the values in Hooke's Law equation to find the force constant:
14.7 N = -k(0.1 m)
k = -14.7 N / (0.1 m)
k ≈ 147 N/m (expressed in positive value since k represents the force constant)
Therefore, the force constant of the spring is approximately 147 N/m.
B. Maximum restoring force acting on the vibrating body:
In Simple Harmonic Motion, the restoring force is maximum at the extremities (maximum displacement). So, when the mass of 4 kg is set into vibration with an amplitude of 12 cm (0.12 m), the maximum restoring force is given by F = kx, where x is the maximum displacement (0.12 m) and k is the force constant (147 N/m).
F = (147 N/m)(0.12 m) = 17.64 N
Therefore, the maximum restoring force acting on the vibrating body is 17.64 N.
C. Period of vibration:
The period of vibration is the time taken for one complete cycle of motion. It can be calculated using the formula T = 2π√(m/k), where m is the mass and k is the force constant.
T = 2π√(4 kg / 147 N/m)
T ≈ 2.787 s
Therefore, the period of vibration is approximately 2.787 seconds.
D. Maximum velocity and maximum acceleration of the vibrating object:
In Simple Harmonic Motion, the maximum velocity and maximum acceleration occur at the equilibrium position (when displacement is zero).
Maximum velocity (v) can be calculated using the formula v = ωA, where ω is the angular frequency and A is the amplitude.
ω = √(k/m) ≈ √(147 N/m / 4 kg) ≈ 3.03 rad/s
v = (3.03 rad/s)(0.12 m) ≈ 0.3648 m/s
Therefore, the maximum velocity of the vibrating object is approximately 0.3648 m/s.
Maximum acceleration (a) can be calculated using the formula a = ω^2A.
a = (3.03 rad/s)^2(0.12 m) ≈ 1.103 m/s^2
Therefore, the maximum acceleration of the vibrating object is approximately 1.103 m/s^2.
E. Speed and acceleration when the displacement is 9 cm (0.09 m):
To find the speed (v) and acceleration (a) when the displacement is 9 cm (0.09 m), we need to use the equations for Simple Harmonic Motion.
v = ω√(A^2 - x^2)
a = ω^2x
Plugging in the values, we get:
x = 0.09 m
A = 0.12 m
ω ≈ 3.03 rad/s
v = (3.03 rad/s)√((0.12 m)^2 - (0.09 m)^2) ≈ 0.1524 m/s
a = (3.03 rad/s)^2(0.09 m) ≈ 0.8217 m/s^2
Therefore, the speed when the displacement is 0.09 m is approximately 0.1524 m/s, and the acceleration is approximately 0.8217 m/s^2.