T = 2 ? / ? = 2 ? /(3.16 rad/s) = s x (0) = A cos ? = A x = (0.0592 m) cos(3.16

Question

T = 2?/? = 2?/(3.16 rad/s) =  s

x(0) = A cos ? = A

x = (0.0592 m) cos(3.16t + 0.18?)
v = -(0.187 m/s) sin(3.16t + 0.18?)
a = -(0.59 m/s2) cos(3.16t + 0.18?)

(a) The first time (greater than zero) that the position is at its maximum value.
  s
(b) The first time (greater than zero) that the velocity is at its maximum value.
  s

Example 15.3 A Block-Spring System Problem A 300 g block connected to a light spring for which the force constant is 3.00 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 5.00 cm from equilibrium and released from rest, as in Figure 15.7.
Find the period of its motion.
Determine the maximum speed of the block.
What is the maximum acceleration of the block?
Express the position, speed, and acceleration as functions of time.
Figure 15.7 A block-spring system that begins its motion from rest with the block at x = A at t = 0. In this case, ? = 0 and thus x = A cos ?t. Solution From Equations 15.9 and 15.10, we know that the angular frequency of a block-spring system is the following.

= 3.16 rad/s

What is the period?

T = 2?/? = 2?/(3.16 rad/s) =  s

We use Equation 15.17.

vmax = ?A = (3.16 rad/s)(5.00 10-2 m) =  m/s

We use Equation 15.18.

amax = ?2A = (3.16 rad/s)2(5.00 10-2 m) =  m/s2

We find the phase constant from the initial condition that x = A at t = 0.

x(0) = A cos ? = A

This tells us that ? = 0. Thus, our solution is x = A cos ?t. Using this expression and the results from (a), (b), and (c), we find the following. (Use t for t.) x = A cos ?t = (  m ) cos( ) v = -?A sin ?t = (  m/s ) sin( ) a = -?2A cos ?t = (  m/s2 ) cos( ) What If? What if the block is released from the same initial position, xi = 5.00 cm, but with an initial velocity ofvi = -0.100 m/s? Which parts of the solution change and what are the new answers for those that do change?

Answer Part (a) does not change-the period is independent of how the oscillator is set into motion. Parts (b), (c), and (d) will change. We begin by considering position and velocity expressions for the initial conditions. (1) x(0) = A cos ? = xi (2) v(0) = -?A sin ? = vi Dividing Equation (2) by Equation (1) gives us the phase constant. = tan ? = ? = 0.18? Now, Equation (1) allows us to find A.

= 0.0592 m

The new maximum speed is the following.

vmax = ?A = (3.16 rad/s)(5.92 10-2 m) = 0.187 m/s

The new magnitude of the maximum acceleration is the following.

amax = ?2A = (3.16 rad/s)2(5.92 10-2 m) = 0.59 m/s2

The new expressions for position, velocity, and acceleration are the following.

x = (0.0592 m) cos(3.16t + 0.18?)
v = -(0.187 m/s) sin(3.16t + 0.18?)
a = -(0.59 m/s2) cos(3.16t + 0.18?)

Explanation / Answer

15.3)

here ,

mass of block , m = 0.300 Kg

force constant , k = 3 N/m

A) period of oscillation = 2pi/sqrt(k/m)

period of oscillation = 2pi/sqrt(3/0.3)

period of oscillation = 1.99 s

the period of oscillation is 1.99 s

B)

angular freuqnecy , w = 2pi/T

w = 6.282/1.99

w = 3.16 rad/s

Now , maximum speed of the block = A*w

maximum speed of the block = 0.05 * 3.16

maximum speed of the block = 0.156 m/s

the maximum speed of the block is 0.156 m/s

C)

maximum acceleration = A*w^2

maximum acceleration = 0.05 * 3.16^2

maximum acceleration = 0.5 m/s^2

the maximum acceleration of the block is 0.5 m/s^2

D)

x = 0.05 * cos(3.16 * t)

velocity , v = -0.156 * sin(3.16 *t)

acceleration , a = -0.5 * sin(3.16 * t)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.