A steel strip, clamped at one end, vibrates with a frequency of 20 Hz and a amplitude of 5 mm at the free end, where a small mass of 2 g is positioned. Find
(i) the velocity of the end when passing through the zero position,
(ii) the acceleration at maximum displacement,
(iii) the maximum kinetic energy of the mass.
Pls solve this for me
Omosewa
Alright, hold on tight while I clown around with these physics questions!
(i) To find the velocity of the end when passing through the zero position, we can use the equation v = ωA, where v is the velocity, ω is the angular frequency, and A is the amplitude. Since the frequency is given as 20 Hz, the angular frequency ω is 2πf, where f is the frequency. Plugging in the values, we have ω = 2π(20) = 40π rad/s. The amplitude A is given as 5 mm, which is 0.005 m. Using the formula, we get v = (40π)(0.005) ≈ 0.628 m/s.
(ii) Now, let's find the acceleration at maximum displacement. The maximum displacement is the same as the amplitude, which is 5 mm or 0.005 m. The acceleration at maximum displacement is given by the equation a_max = ω^2A. Plugging in the values, we get a_max = (40π)^2(0.005) ≈ 125.663 m/s^2. So, the acceleration at maximum displacement is approximately 125.663 m/s^2.
(iii) Finally, let's calculate the maximum kinetic energy of the mass. The maximum kinetic energy (K.E_max) is given by the equation K.E_max = 0.5mv^2, where m is the mass and v is the velocity. The mass is given as 2 g, which is 0.002 kg. We've already calculated the velocity at the zero position as 0.628 m/s. Plugging in the values, we have K.E_max = 0.5(0.002)(0.628)^2 ≈ 0.000394 J.
So, there you have it! The answers to your physics questions with a pinch of clowning around.
To solve these problems, we can use the equations for simple harmonic motion.
(i) To find the velocity of the end when passing through the zero position, we can use the equation for velocity in simple harmonic motion:
v = ω * A
where v is the velocity, ω is the angular frequency, and A is the amplitude.
Given that the frequency is 20 Hz, we can find the angular frequency using the formula:
ω = 2πf
Substituting the values:
ω = 2π * 20 Hz = 40π rad/s
Given that the amplitude is 5 mm, we convert it to meters:
A = 5 mm = 0.005 meters
Now we can calculate the velocity:
v = ω * A = 40π rad/s * 0.005 meters = 0.2π m/s (approximately 0.628 m/s)
So, the velocity of the end when passing through the zero position is approximately 0.628 m/s.
(ii) To find the acceleration at maximum displacement, we can use the equation for acceleration in simple harmonic motion:
a = -ω^2 * x
where a is the acceleration, ω is the angular frequency, and x is the displacement.
Given that the maximum displacement is the amplitude A, we substitute the values:
a = -ω^2 * A = - (40π rad/s)^2 * 0.005 meters
= - (1600π^2) m/s^2
So, the acceleration at maximum displacement is - (1600π^2) m/s^2.
(iii) To find the maximum kinetic energy of the mass, we can use the equation for kinetic energy in simple harmonic motion:
K.E. = (1/2) * m * ω^2 * A^2
where K.E. is the kinetic energy, m is the mass, ω is the angular frequency, and A is the amplitude.
Given that the mass is 2 g, we convert it to kilograms:
m = 2 g = 0.002 kg
Substituting the values:
K.E. = (1/2) * 0.002 kg * (40π rad/s)^2 * (0.005 meters)^2
= 0.00005π^2 J
So, the maximum kinetic energy of the mass is approximately 0.00005π^2 J.
To find the velocity of the end when passing through the zero position, we need to consider the relationship between velocity, frequency, and amplitude in simple harmonic motion.
The equation for the displacement of an object undergoing simple harmonic motion is given by:
x = A * cos(ωt)
Where:
- x is the displacement of the object at time t
- A is the amplitude of the motion
- ω is the angular frequency, which is given by ω = 2πf, where f is the frequency of the motion
To find the velocity when passing through the zero position, we need to differentiate the displacement equation with respect to time:
v = dx/dt = -A * ω * sin(ωt)
- Differentiating cos(ωt) with respect to time gives -ω * sin(ωt)
Now, we can substitute the values:
Amplitude (A) = 5 mm = 0.005 m
Frequency (f) = 20 Hz
Angular frequency (ω) = 2πf = 2π * 20 = 40π rad/s
Using the equation for velocity, we can calculate the velocity at the zero position:
v = -A * ω * sin(ωt) = -0.005 * 40π * sin(0) = 0
Therefore, the velocity of the end when passing through the zero position is zero.
Now, let's move on to finding the acceleration at maximum displacement.
The equation for acceleration in simple harmonic motion is the second derivative of the displacement equation:
a = d²x/dt² = -A * ω² * cos(ωt)
- Differentiating -ω * sin(ωt) with respect to time gives -ω² * cos(ωt)
Using the expression for acceleration, we can calculate the acceleration at maximum displacement:
a = -A * ω² * cos(ωt) = -0.005 * (40π)² * cos(π/2) = -800π² m/s²
Therefore, the acceleration at maximum displacement is -800π² m/s².
Lastly, let's determine the maximum kinetic energy of the mass.
The kinetic energy of an object undergoing simple harmonic motion is given by the formula:
K.E = (1/2) * m * v²
Where:
- K.E is the kinetic energy
- m is the mass of the object
- v is the velocity of the object
The mass of the object (m) is given as 2 g = 0.002 kg.
Using the equation for velocity, we found that the velocity of the end when passing through the zero position is 0 m/s. Therefore, the kinetic energy at this point would be:
K.E = (1/2) * m * v² = (1/2) * 0.002 * 0² = 0 J (Joules)
Hence, the maximum kinetic energy of the mass is zero.