Question

2)A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.2 cm. (Assume the position of the object is at the origin at t = 0.)

(a) Calculate the maximum value of its speed.
Answer must be in cm/s

(b) Calculate the maximum value of its acceleration.
Answer must be in cm/s²

(c) Calculate the value of its speed when the object is 8.20 cm from the equilibrium position.
Answer must be in cm/s

(d) Calculate the value of its acceleration when the object is 8.20 cm from the equilibrium position.
Answer must be in cm/s²

(e) Calculate the time interval required for the object to move from x = 0 to x = 2.20 cm.
Answer must be in s

5) Calculate the length of a pipe that has a fundamental frequency of 476 Hz. (Take the speed of sound in air to be 343 m/s.)

(a) Assume the pipe is closed at one end.
Answer must be in m

(b) Assume the pipe is open at both ends.

Answer must be in m

Answer 1

First question -

(2) Force constant of the spring, k = 8.00 N/m

Mass of the object, m = 0.520 kg

Therefore,

ω = √(k/m) = √(8.00N/m / 0.520kg) = 3.92 rad/s

Given that, amplitude of the motion = 10.2 cm

From the above data, write the equation for simple harmonic motion -

x(t) = 10.2cm * sin(3.92*t) -------------------------------------------(i)

(a) Maximum speed = A*ω = 10.2 * 3.92 cm/s = 39.98 cm/s (Answer)

(b) Maximum acceleration = A*ω²

= 10.2 * 3.92²

= 156.7 cm/s²

(c) Total Mechanical Energy (TME) = ½kA² = ½ * 8.00N/m * (0.102m)² = 0.0416 J

When x = 0.082 m, PE = ½ * 8.00N/m * (0.082m)² = 0.0269 J

leaving KE = 0.0416 - 0.0269 = 0.0147 J = ½ * 0.520kg * v²

=> v² = (2 x 0.0147) / 0.520 = 0.0565

=> v = 0.238 m/s = 23.8 cm/s (Answer)

(d) Acceleration, a = kx / m

= (8.00N/m * 0.082m) / 0.520kg

= 1.26 m/s² = 126.0 cm/s² (Answer)

(e) From equation (i) -

2.20 cm = 10.2cm * sin(3.92*t)

3.92*t = arcsin(2.20/10.2) = 0.214

=> t = 0.214 / 3.92 = 0.0546 s (Answer)

I hope you understand the solution.