The tip of one prong of a tuning fork undergoes SHM of frequency L000 Hz and amplitude 0.40 mm. For this tip, what is the magnitude of the (a) maximum acceleration, (b) maxi- mum velocity, (c) acceleration at tip displacement 0.20 ffiffi, and (d) velocity at tip displacement 0.20 mm

The tip of one prong of a tuning fork undergoes SHM of frequency 1000 Hz and amplitude 0.40 mm. For this tip, what is the magnitude of the

(a) maximum acceleration,

(b) maximum velocity,

(c) acceleration at tip displacement 0.20 mm, and

(d) velocity at tip displacement 0.20 mm?

(a) 15795.5m/s²

(b) 2.5m/s

(c) 7897.7 m/s²

(d) 2.2m/s

The displacement, y, of a body undergoing simple harmonic motion (SHM) is given by

y = A sin (ωt + φ)             ------------------(i)

Where;

A = maximum displacement or amplitude of the body

ω = angular frequency of the body

t = time taken for the displacement

φ = phase constant

The velocity, v, of the body can be found by differentiating equation (i) as follows;

v = Aω cos (ωt + φ)          ------------------(ii)

Where;

Aω = maximum velocity or amplitude of the velocity of the body

Also, the acceleration of the body can be found by differentiating equation (ii) as follows;

a = -Aω² sin(ωt + φ)                  --------------------(iii)

Where;

-Aω² = maximum acceleration or amplitude of the acceleration of the body

(a) From equation (iii), the magnitude of the maximum acceleration is given by;

= Aω²           ----------------(iv)

Where;

A = amplitude = 0.40mm = 0.00040m

ω = 2 π f            [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]

=> ω = 2 x 3.142 x 1000 = 6284 rad/s

Substitute these values into equation (iv) as follows;

= 0.00040 x 6284² = 15795.5m/s²

Therefore, the magnitude of the maximum acceleration is 15795.5m/s²

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(b) From equation (ii), the magnitude of the maximum velocity , is given by;

= Aω          ----------------(v)

Where;

A = amplitude = 0.40mm = 0.00040m

ω = 2 π f            [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]

=> ω = 2 x 3.142 x 1000 = 6284 rad/s

Substitute these values into equation (v) as follows;

= 0.00040 x 6284 = 15795.5m/s²

Therefore, the magnitude of the maximum velocity is 2.5m/s

========================================================

(c) Comparing equations (i) and (iii),  equation (iii) can be written as;

a = -ω² y            -------------------(vi)

Therefore, to get the acceleration at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m and ω = 6284rad/s into equation (vi) as follows;

a = - 6284² x 0.00020

a = - 7897.7 m/s²

Therefore, the magnitude of the acceleration at the tip displacement is 7897.7 m/s²

========================================================

(d) Recall that;

sin²θ + cos²θ = 1

=> cos²θ = 1 - sin²θ

=> cosθ = √(1 - sin²θ )

=> cos (ωt + φ)  = √(1 - sin² (ωt + φ))

Substitute this value into equation (ii) as follows;

v = Aω √(1 - sin² (ωt + φ))

v = ω√(A² - A²sin² (ωt + φ))              

Now, comparing the equation above and equation (i), the equation above can be written as;

v = ω√(A² - y²)           -------------(vii)

Therefore, to get the velocity at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m, ω = 6284rad/s and A = 0.00040m into equation (vii) as follows;

v = 6284√(0.00040² - 0.00020²)

v = 6284√(0.00000060)

v = 2.2m/s

Therefore, the magnitude of the velocity at the tip displacement is 2.2 m/s

graphing motion: mastery test

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