A mass hangs from a spring and is set into vertical oscillation with a frequency

Question

A mass hangs from a spring and is set into vertical oscillation with a frequency of 1.25 Hz. When the mass is at the highest point of its motion, the string is stretched by 4.00cm. A stopwatch is started during the motion and it is found that the mass is 5.00cm below the equilibrium position and is moving upwards when t=0.200s. Calculate:
a) the angular frequency
b) the extension of the spring when the mass is at the equilibrium position
c) the amplitude of the motion
d) the phase angle
e) the speed of the mass when t=0.200s.

Explanation / Answer

let m = mass of the oscillating object
find k of the spring:
= 2pif = (k/m)
2pi*1.25 = (k/m)
(2pi*1.25)^2 = k/m
k = 61.685m

and F = x*k with x = extension of the spring
x = F/k with F = mg and k = 61.685m
x = mg/(61.685m)
x = g/61.685
x = 0.119 meter
this x is the distance between the unloaded spring and the spring with the mass attached.
Since this position is also the equilibrium point about which the mass oscillates, the x is
the amplitude + the 4 cm that the distance x is larger than the amplitude.
It follows that the amplitude is 0.119 m - 0.04 m = 0.079 m (= 7.9 cm).

Now you can set up the equation of the oscillation:
y = A*sin(t + )
y = 0.079*sin((2pif)*t + ) with 2pif = 7.854
y = 0.079*sin(7.854t + ) --> first derivative = v
v = 0.079*7.854*cos(7.854t + )

I can not answer d and e, because one has to know where the mass is at t = 0.
i.e., from what point the oscillation starts at t=0. If you do not know this, t = 0.2 s can be any time.
you can try to put t = 0.2 into the equation for the position and the velocity to find :
- 0.05 = 0.079sin(7.854*0.2 + ), and then put this into the equation for v:
v = 0.079*7.854cos(7.854*0.2 + ).

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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