Question

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 2.5 cm and released. The ball makes 23 oscillations in 22 s seconds.

a) What is its mass of the ball? (in g)

b) What is its maximum speed? (in cm/s)

Answer 1

Spring constant k=15 N/m Amplitude of oscillation: The distance of pull down is equal to the amplitude of the oscillation. A =2.5 cm =2.5 cm 0.01 m 31 cm = 0.025 m Time period of oscillation: It is the ratio of time taken (t) and number of oscillations Izt 23 Angular frequency:

Part (a): Calculate the mass of the ball: The expression for spring constant in terms of mass (m) and angular frequency is given below. k = mo? Rewrite the above equation for mass. 2377 rad s in the above equation. Substitute k = 15 N/m and @=> 15 N/m m =- = 0.3476 kg (1000 g =0.3476 kg 1 kg) = 347.6 g Part (b): Calculate the maximum speed of the ball. The expression for maximum speed is, VaR = 04 Substitute o=* rad/sand A=0.025 m in the above equation. Vax = ( 237 rad's (0.025 m) =0.1642 ms ( 100.com =164.2 cm/s

Answer 2

periode of this spring is : 20/40 = 0.5 second
to determine mass:

T=2*pi*V(m/k)
0.5 = 2*pi*V(m/15)
we can get m= 0.095 kg

maximum speed, we can use :
EP1+EK1=EP2+EK2
0.5*k*x^2 + 0 =0.5*m*v^2 +0
0.5*15*0.06^2 = 0.5*0.095*v^2
0.027=0.0475*v^2
v^2=0.568
v=0.75 m/s

Answer 3

k = 15.8 N/m
A = 0.075 m
? = 39(2p) = 78p
t = 20 s


Part A.

? = ?/t = 78p/20

? = 3.9p

k = m?²

15.8 = m(3.9p)²

m = 0.105 kg


Part B:

maximum speed is

V = A? = 0.075( 3.9p)

V = 0.919 m/s