The figure below gives the position of a 22 g block oscillating in SHM on the en

Question

The figure below gives the position of a 22 g block oscillating in SHM on the end of a spring. (The horizontal axis is marked in increments of 30 ms.)

(a) What is the maximum kinetic energy of the block?
?J
(b) What is the number of times per second that maximum is reached?
?times per second

(c) For the quantities listed below, please give:
· its maximum magnitude
· the time it is first reached after t=0 s
(Note: Be wary of the limitations of a purely graphical approach.)

vmax = ? m/s
t = ?ms

amax = ? m/s2
t = ?ms

Explanation / Answer

Given,

mass = m = 22g = 0.022 kg ;

(a)The maximum kinetic of the block will be equal to the stored potential energy of the spring. So,

KE(max) = PE(spring)

KE(max) = 1/2 k x2

we know that, T = 2 pi sqrt (m/k), this will give us:

k = m x 4 x pi2 / T2

The period of one complete oscillation from figure = T = 120 ms = 0.12 s

k = 0.022 x 4 x 9.86 / 0.12 x 0.12 = 60.26 N/m

x(max) = 7 cm = 0.07 m

KE(max) = 1/2 x 60.26 x (0.07)2 = 0.147 J

Hence, KE(max) = 0.147 J

(b)We need to determine frequency here.

We know that,

f = 1/T = 1/0.12 = 8.33 per sec

Hence, f = 8.33 per sec

(c) In part a we have calculated the max KE so

KEmax = 1/2 m v(max)2 = 0.147 => v = sqrt ( 2 x 0.147 / 0.022)

v(max) = 3.66 m/s

at, t= 0.030 s

a(max) = v/t = 3.66 m/s / 0.03 = 122 m/s2

a(max) = 122 m/s2

at t = 0.03 s

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