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**Question 1:**

An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0.00 s. It then oscillates with a period of 11.4 s and a maximum speed of 52.9 cm/s. What is the glider's position at t=0.561 s?

==> ANSWER:

T = 11.4s, Vmax = 52.9 cm/s

I found out W (angular frequency) to = 0.175 pi = 0.55498 rad/s

I also found A = 96.22 cm I used the equation

X(t) = A cos ( w t )

X(0.561) = (96.22) cos ( 0.55498 x 0.561 )

= 96.2186 cm

apparently this is the wrong answer....idk wat i am doing wrong though..any pointers?

**Question 2:**

A 543.0 g mass oscillates with an amplitude of 12.4 cm on a spring whose spring constant is 27.3 N/m. At t=0.00 s the mass is 4.36 cm to the right of the equilibrium position and moving to the right. Determine the phase constant. Use a cosine function to describe the simple harmonic motion.

==> ANSWER:

Knowns: Xinitial = 4.36 cm A= 12.4 cm

phase constant = cos^-1 ( Xinitial / A )

= cos^-1 (4.36 / 12.4)

= 69.41 rad.....i think the units are radian rite?

Again i got the wrong answer and idk where i went wrong....

**Question 3:**

An object in simple harmonic motion oscillates with a period of 4.00 s and an amplitude of 9.79 cm. How long does the object take to move from x=0.00 cm to x=6.50 cm?

==> ANSWER:

Knowns: T=4s A= 9.78m

I found: w= 2pi/ T = 0.3927 rad/s

Vmax = wA = 3.845 cm/s

i have no idea how else to go about this question....any suggestions?

**Question 4:**

The velocity of an object in simple harmonic motion is given by v(t)= -(0.226 m/s)sin(15.0t + 2.00π), where t is in seconds. What is the first time after t=0.00 s at which the velocity is -0.110 m/s?

==> ANSWER: so since the equation for velocity is

V(t) = -(Vmax) sin ( wt + phase constant )

we know that,

Vmax = 0.226m/s W= 15 rad/s phase constant= 2pi rad

i figured that the speed would be the same after a period....so i set out to find the period,

T = 2pi / w = 0.4189s

however that is the wrong answer, is there a flaw in my logic?

**Question 5:**

A 6.00 g spider is dangling at the end of a silk thread. You can make the spider bounce up and down on the thread by tapping lightly on his feet with a pencil. You soon discover that you can give the spider the largest amplitude on his little bungee cord if you tap exactly once every 7.00 s seconds. What is the spring constant of the silk thread.

==> ANSWER:

m= 6g = 0.006 kg

T= 7s

k = ?

I used the equation T = 2pi ( m / k )^(1/2)

i rearranged it for k = (m 4pi) / T^2 = 1.54 x 10^-3 N/m

again this was the wrong answer....plz help.

Sry for posting so many questions but i didnt want to start many threads and the topic is all related its just that im studying for a test and these are the questions im stuck on....any help would be greatly appreciated..THANKS AGAIN!