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A 200g particle oscillating in SHM travels 66cm between the two extreme points in its motion with an average speed of 110cm/s. Find: a) The angular frequency. s−1 b) The maximum force on the particle. N b) The maximum speed.

User Kevinsss
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Final answer:

The angular frequency of the particle in SHM is approximately 10.49 rad/s. The maximum force on the particle is approximately -2.91 N. The maximum speed of the particle is approximately 6.91 m/s.

Step-by-step explanation:

In simple harmonic motion (SHM), the angular frequency, denoted by ω, is given by the formula:

ω = 2πf

where f is the frequency of the motion. In this case, the average speed of the particle is given as 110 cm/s. The distance traveled between the two extreme points is 66 cm. Since the average speed is equal to the displacement divided by the time taken, we can calculate the time taken as 66 cm / 110 cm/s = 0.6 s. Therefore, the frequency is 1 / 0.6 s = 1.67 Hz. Using the formula for angular frequency, we can find:

ω = 2πf = 2π * 1.67 Hz ≈ 10.49 rad/s

The maximum force on the particle can be calculated using the formula for simple harmonic motion:

F = -kx

where F is the maximum force, k is the force constant, and x is the amplitude of the motion. The force constant can be found using the formula:

k = mω²

where m is the mass of the particle. In this case, the mass is given as 200 g, which is equal to 0.2 kg. Substituting the values into the formula, we have:

k = 0.2 kg * (10.49 rad/s)² ≈ 4.42 N/m

Substituting the values of k and x (66 cm = 0.66 m) into the formula for maximum force, we have:

F = -(4.42 N/m)(0.66 m) ≈ -2.91 N

The maximum speed of the particle can be found using the formula:

vmax = ωA

where vmax is the maximum velocity and A is the amplitude. Substituting the values of ω (10.49 rad/s) and A (0.66 m), we have:

vmax = (10.49 rad/s)(0.66 m) ≈ 6.91 m/s

User Dargaud
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