Question

Im stuck on questions (

Questions 2a-c (refer to a 0.250 kg pendulum, and then answer 2a-d)

An object suspended from a spring with a spring constant of 2.56 N/m vibrates with a frequency of 0.148 Hz. What is the mass of the object?

f=½(pi) x sqroot(k/m) 0.148Hz=½(pi) x sqroot(2.56N/m / m) 0.148 x 2(pi)= sqroot(2.56)/m 0.93 = sqroot(2.56Hz/m) 0.93^2=sqroot(2.56Hz/m) m=2.66/0.93^2 = 2.96kg The mass is 2.96kg

What is the acceleration of the object at a displacement of -0.120 m from the equilibrium position?

a=-wx sqroot(k/m)=sqroot(2.56/2.96=0.93rad/s^2) a=-0.93x(-0.12)=0.112m/s^2 The acceleration is 0.112m/s^2

Questions 2a-c refer to a 0.250 kg pendulum. What length of the pendulum would be needed to oscillate at the same frequency as the object in question 1?

What would be the restoring force on the pendulum at an angle of 6.24° from the equilibrium position?

The pendulum is pulled aside until it is 0.386 m above its lowest position and released. The pendulum is designed to emit sound waves at a frequency of 440 Hz; however, as it swings toward and away from an observer, the frequency appears to vary slightly. What would be the apparent frequency of the sound from the pendulum as it swings at its maximum speed toward an observer? Assume the speed of sound is 345 m/s.

Answer 1

Time period of a pendulum is given by .------(i)

Therefore, the frequency is -----(ii)

given the frquency is .

1) using the value of frequency in (ii), we can get the value of l :

[answer]

2) m = mass = 0.25 kg

the restoreing force on a pendulum is component of gravity along tangent.

therfore, restoreing force = [theta is the angle made by the length with lower vertical]

= [answer]

3) maximum speed of the pendulum is when it is at lowest position, and it is equal to u =

now, the source is moving towards stationary observer with speed u = 2.752 m/s

speed of source = vs = 2.752 m/s

speed of sound = v = 345 m/s

f = 440 Hz

the apparent frequncy observed by observer = [answer]