In which type of bone does ossification occur in the membranes?

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in

navik [9.2K]

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) = $\frac{1}{2π}.\sqrt{\frac{k}{m} }$

inserting the value of spring constant (k) as 98m into the equation above

f = $\frac{1}{2π}.\sqrt{\frac{98m}{m} }$

f = $\frac{1}{2π}.\sqrt{98}$

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

( $0.5ky^{2} - mgy) + 0.5mv^{2}$

v = $\sqrt{2gy - \frac{k}{m}.y^{2}}$

where y = position of the spring = 8 cm (0.08m) and k = 98m as in (A) above

v = $\sqrt{2 x 9.8 x 0.08 - \frac{98m}{m}.0.08^{2}}$

v = $\sqrt{1.568 - (98 x 0.0064}$

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

$\frac{1}{2π}.\sqrt{\frac{k}{m + 300} }$ = 0.5 x $\frac{1}{2π}.\sqrt{\frac{k}{m} }$

$\sqrt{\frac{k}{m + 300} }$ = 0.5 x $\sqrt{\frac{k}{m} }$

$(\sqrt{\frac{k}{m + 300} })^{2} =(0.5 x \sqrt{\frac{k}{m} })^{2}$

$\frac{k}{m + 300} =0.25 x \frac{k}{m}$

$\frac{1}{m + 300} =0.25 x \frac{1}{m}$

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 = $\frac{(m+0.3)g}{k}$

recall that k = 98m

x = $\frac{(m+0.3)g}{98m}$

now substituting the values of m and g into the equation

x = $\frac{(0.1 + 0.3) x 9.8}{98 x 0.1}$

x = 0.4 m

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