Answer:
(A) 1.58 Hz
(B) 0.99 m/s
(C) 0.1 kg
(D) 0.4 m
Explanation:
extension of the spring (x) = 10 cm = 0.1 m
acceleration due to gravity (g) = 9.8 m/s^{2}
(A) force = mg = kx
where m = mass
g = acceleration due to gravity
x = extension
mg = kx
substituting the values 0f g and x we have
9.8m = 0.1k
therefore k = 9.8m/0.1
k = 98m
formula for frequency (f) =
inserting the value of spring constant (k) as 98m into the equation above
f =
f =
f = 1.58 Hz
(B) find the speed of the object when it is 8 cm below its initial position
from the conservation of energy,
initial potential energy (U) + kinetic energy (K.E) = 0
(
v =
where y = position of the spring = 8 cm (0.08m) and k = 98m as in (A) above
v =
v =
v = 0.99 m/s
(C) find the mass (m) of the object when an object of mass 300 g is attached to the first object, after which the system oscillates with half the original frequency.
after addition of the 300 g mass
new frequency = half the initial frequency
= 0.5 x
= 0.5 x
0.25 (m + 300) = m
m - 0.25m = 0.25 x 300
m- 0.25m = 75
0.75 m = 75
m = 100 g = 0.1 kg
(D) find the new equilibrium position
from mg = kx we can find the new equilibrium position (x)
where m = m + 300 = m + 0.3 (in kg)
(m+0.3)g = kx
x =
recall that k = 98m
x =
now substituting the values of m and g into the equation
x =
x = 0.4 m