1. A particle moving with simple harmonic motion has maximum displacement of 50 cm angular velocity of 1.02rad/s. Calculate the (a) the maximum velocity (b) maximum acceleration of the particle (c) the speed and acceleration of the particle when it is 30cm from the center of the oscillation (d) the period
2. A 500g car moving with a velocity of 20cm/s collides with utility pole brought to rest in 0.3sec .
Find the magnitude of force exerted on the car during collision?
To answer these questions, we will use the equations related to simple harmonic motion and Newton's second law of motion.
1. (a) Maximum Velocity:
The maximum velocity of a particle in simple harmonic motion is given by the equation:
v_max = ω * A
where v_max is the maximum velocity, ω is the angular velocity, and A is the maximum displacement.
Substituting the given values:
v_max = 1.02 rad/s * 50 cm = 51 cm/s
(b) Maximum Acceleration:
The maximum acceleration of a particle in simple harmonic motion is given by the equation:
a_max = ω^2 * A
where a_max is the maximum acceleration, ω is the angular velocity, and A is the maximum displacement.
Substituting the given values:
a_max = (1.02 rad/s)^2 * 50 cm = 52.04 cm/s^2
(c) Speed and Acceleration at 30 cm:
To find the speed and acceleration when the particle is at 30 cm from the center of oscillation, we need to use the equations for velocity and acceleration in simple harmonic motion. The equations are as follows:
v = ω * √(A^2 - x^2)
where v is the speed of the particle and x is the displacement from the center.
a = ω^2 * x
where a is the acceleration and x is the displacement from the center.
Substituting the given values:
For speed (v) when x = 30 cm:
v = 1.02 rad/s * √(50^2 - 30^2) cm = 41.46 cm/s
For acceleration (a) when x = 30 cm:
a = (1.02 rad/s)^2 * 30 cm = 31.08 cm/s^2
(d) Period:
The period of a particle in simple harmonic motion is the time taken for one complete oscillation. It is given by the equation:
T = 2π / ω
where T is the period and ω is the angular velocity.
Substituting the given value:
T = 2π / 1.02 rad/s ≈ 6.16 s
2. To calculate the magnitude of force exerted on the car during the collision, we will use Newton's second law of motion, which states that force (F) is equal to the mass (m) multiplied by acceleration (a):
F = m * a
Given:
Mass of the car (m) = 500 g = 0.5 kg
Time taken for the car to come to rest (t) = 0.3 s
Initial velocity of the car (u) = 20 cm/s = 0.2 m/s
Final velocity of the car (v) = 0 m/s (car comes to rest)
To calculate acceleration (a), we can use the equation:
a = (v - u) / t
Substituting the values:
a = (0 m/s - 0.2 m/s) / 0.3 s = -0.67 m/s^2 (negative sign indicates deceleration)
Now, substituting the values in the equation for force (F):
F = 0.5 kg * -0.67 m/s^2 = -0.335 N (magnitude of force cannot be negative, so taking the absolute value)
Hence, the magnitude of the force exerted on the car during the collision is approximately 0.335 N.
1.
(a) To calculate the maximum velocity, we can use the formula: vmax = ω * A, where vmax is the maximum velocity, ω is the angular velocity, and A is the maximum displacement.
Given:
A = 50 cm
ω = 1.02 rad/s
Substituting these values:
vmax = 1.02 rad/s * 50 cm = 51 cm/s
Therefore, the maximum velocity of the particle is 51 cm/s.
(b) To calculate the maximum acceleration, we can use the formula: amax = ω^2 * A, where amax is the maximum acceleration.
Given:
A = 50 cm
ω = 1.02 rad/s
Substituting these values:
amax = (1.02 rad/s)^2 * 50 cm = 52.02 cm/s^2
Therefore, the maximum acceleration of the particle is 52.02 cm/s^2.
(c) To calculate the speed and acceleration of the particle when it is 30 cm from the center of oscillation, we can use the formulas: v = ω * x and a = ω^2 * x, where v is the velocity, a is the acceleration, ω is the angular velocity, and x is the displacement.
Given:
x = 30 cm
ω = 1.02 rad/s
Substituting these values:
v = 1.02 rad/s * 30 cm = 30.6 cm/s
a = (1.02 rad/s)^2 * 30 cm = 30.924 cm/s^2
Therefore, when the particle is 30 cm from the center of oscillation, its speed is 30.6 cm/s and its acceleration is 30.924 cm/s^2.
(d) To calculate the period, we can use the formula: T = 2π/ω, where T is the period and ω is the angular velocity.
Given:
ω = 1.02 rad/s
Substituting this value:
T = 2π / 1.02 rad/s ≈ 6.19 s
Therefore, the period of the simple harmonic motion is approximately 6.19 seconds.
2.
To find the magnitude of force exerted on the car during the collision, we can use Newton's second law of motion, which states that force is equal to mass times acceleration (F = m * a).
Given:
m (mass of the car) = 500 g = 0.5 kg
v (initial velocity of the car) = 20 cm/s = 0.2 m/s
t (time taken to come to rest) = 0.3 s
First, we need to find the acceleration of the car. Using the formula v = u + at, where u is the initial velocity, v is the final velocity (which is 0 in this case), and t is the time taken:
0 = 0.2 m/s + a * 0.3 s
Solving for a:
a = -0.2 m/s / 0.3 s = -0.67 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.
Now, we can find the force exerted using the formula F = m * a:
F = 0.5 kg * (-0.67 m/s^2) = -0.335 N
Therefore, the magnitude of the force exerted on the car during the collision is approximately 0.335 N.