The lenth of a simple pendelum is 1.85m and the mass of the particle (the \"bob\

Question

The lenth of a simple pendelum is 1.85m and the mass of the particle (the "bob") at the end of the cable is 0.38 kg. The pendelum is pulled away from its equilibrium position until it reaches a height above its equilibrium position of 8.5cm. Assume friction can be neglected and that the resulting oscillatory motion is simple harmonic motion...

a.) What is the frequency (Hz) of the motion?

b.) What is the "bob's" speed as it passes through the lowest point of the swing?

c.) Determine the value of theta at the point of maximum displacement from the equilibrium position.

Explanation / Answer

given

L = 1.85 m

m = 0.38 kg

h = 8.5 cm = 0.085 m

a) we know, Time periode of simple pendulum, T = 2*pi*sqrt(L/g)

= 2*pi*sqrt(1.85/9.8)

= 2.73 s

so, frequency, f = 1/T

= 1/2.73

= 0.37 Hz

b) v = sqrt(2*g*h)

= sqrt(2*9.8*0.085)

= 1.3 m/s

c) h = L*(1 - cos(theta))

h/L = 1 - cos(theta)

cos(theta) = 1 - h/L

cos(theta) = 1 - 0.085/1.85

cos(theta) = 0.954

theta = cos^-1(0.954)

= 17.4 degrees

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.