Question

An air-track glider attached to a spring oscillates between the 12.0 cm mark and the 65.0 cm mark on the track. The glider completes 14.0 oscillations in 34.0 s . What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider

Answer 1

(a) 2.4s

(b) 0.41Hz

(c) 26.5cm

(d) 68.1m/s

Step-by-step explanation:

(a) To calculate the period we can use the formula:

`T=(1)/(f)=(1)/((14.0)/(34.0s))=2.4s`

(b) the frequency is the inverse of period:

`f=(1)/(T)=(1)/(2.4)=0.41Hz`

(c) The amplitude is the max distance to the equilibrium position, that is:

`A=(65.0cm-12.0cm)/(2)=26.5cm`

(d) the maximum speed of the glider can be computed by using:

`v_(max)=\omega A\\\\\omega=2\pi f=2\pi (0.41Hz)=2.57rad/s\\\\v_(max)=(2.57rad/s)(26.5cm)=68.105(cm)/(s)`

hope this helps!!

Answer 2

a) T = 2.429 s

b) f = 0.412 Hz

c) A = 26.5 cm

d) v = 68.6 cm/s

Step-by-step explanation:

a) The period is equal:

T = 34/14 = 2.429 s

b) The frequency is equal:

f = 1/T = 1/2.429 = 0.412 Hz

c) The amplitude is equal:

`A=(62-12)/(2) =26.5cm`

d) The maximum speed of the glider is:

`v=aw=A*2\pi *f=26.5*2*\pi *0.412=68.6cm/s`