1. What is the maximum value of the acceleration a when x = Acos(?t + ?)? a. ? b
ID: 1391928 • Letter: 1
Question
1. What is the maximum value of the acceleration a when x = Acos(?t + ?)?
a. ? b. ?t c. ? 2 A d. A? e. ? A
2. A simple pendulum on the Earth has a period of one second. What would be its period in s on the moon where the acceleration due to gravity is 1/6 that of Earth?
a. 6.00 b. 2.45 c. 1.00 d. 0.408 e. 0.167
3. The lowest A on a piano has a frequency of 27.5 Hz. If the tension in the 2.0 meter string is 308 N, and one-half wavelength occupies the wire, what is the mass of the wire in kg?
a. 0.025 b. 0.051 c. 0.72 d. 0.81 e. 0.37
EXPLAIN HOW YOU GOT THE ANSWER PLEASE
Explanation / Answer
x = A*cos(wt+pi)
speed = v(t) = dx/dt
v(t) = -A*sin(wt+pi)*w = -w*A*sin(wt+pi)
acceleration = a = dv/dt
a(t) = -w*A*os(wt+pi)*w
a(t) = -w^2*A*cos(wt+pi)
a(t) = -w^2*x
acceleration is maximum when x is maximum .
maximum value of x = A
amax = w^2*A <-------answer
option C
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#2)
time period of a simple pendulum T = 2*pi*sqrt(L/g)
Tearth / Tmoon = sqrt(gmoon/gearth)
gmoon = 1/6*gearth
gmoon/gearth = 1/6
1/Tmoon = sqrt( 1/6)
Tmoon = sqrt6
Tmoon = 2.45 s
---------------
3)
wavelength /2 = L
wavelength = 2L = 282 = 4 m
speed = wavelength*f = 4*27.5 = 110 m/s
but speed in a string = v= sqrt(T/u)
u = mass per unit length = m/L
v = sqrt(T*L/m)
m = T*L/v^2
m = (308*2)/(110*110)
mass = m =0.051 kg <------------answer
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