Question

A block of mass 500 g is attached to a horizontal spring, whose force constant is 25. 0 n/m. The block is undergoing simple harmonic motion with an amplitude of 6. 00 cm. At t=0 the block is 4. 00 cm to the left of its equilibrium position and is moving to the right. At what time t1 will it first reach the limit of its motion to the right?

Answer 1

The time taken for the car to reach the limit or amplitude of the motion is 0.33 s.

Equation for the simple harmonic motion of the block

The equation for the simple harmonic motion of the block is calculated as follows;

y = A sin(ωt + Φ)

where;

• A is the amplitude
• ω is the angular speed
• t is the time of motion
• Φ is the phase angle

Angular speed

The angular speed is calculated as follows;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{25}{0.5} } \\\\\omega = 7.07 \ rad/s$

Phase angle

$\Phi = sin^{-1} (\frac{-4}{6} )\\\\\Phi = -0.73 \ rad$

The time taken for the car to reach the limit or amplitude of the motion is calculated as follows;

0.06 = 0.06sin(7.07t - 0.73)

1 = sin(7.07t - 0.73)

7.07t - 0.73 = sin⁻¹(1)

7.07t - 0.73 = 1.57

7.07t = 2.3

t = 2.3/7.07

t = 0.33 s

Learn more about simple harmonic motion here: brainly.com/question/17315536