A mass m = 4.6 kg is attached to a vertical spring with k = 208 N/m and is set i

Question

A mass m = 4.6 kg is attached to a vertical spring with k = 208 N/m and is set into motion. (a) What is the frequency of the oscillation? 1.07 Hz (b) If the amplitude of the oscillation is 3.1 cm, what is the maximum value of the velocity? ________ m/s (c) How long does it take the mass to move from y = 1.5 cm to y = 2.5 cm? ___________ s (d) If the mass is oscillating with a maximum speed of 45 m/s, what is the amplitude? ____________ m (e) If the spring constant is increased by a factor of two and the maximum kinetic energy of the mass is the same, by what factor does the amplitude change? ___________ % of the original amplitude

Explanation / Answer

(a) the frequency of the oscillation = (1/2pi)*sq rt(k/m) = (1/(2*3.14))*sq rt[208/4.6] = 1.0707 Hz

(b) If the amplitude of the oscillation is 3.1 cm, the maximum value of the velocity = 2*pi*f*3.1 cm/s
= 2*3.14*1.07*3.1 = 20.83 cm/s

(c)Let it take, T s for the mass to move from y = 1.5 cm to y = 2.5 cm. We have
y = 3.1*sin[2*pi*ft]
1.5=3.1*sin [2*pi*f*t] gives t={1/(2*pi*f)}arcsin[1.5/3.1]

={1/(2*3.14*1.07)*arcsin[1.5/3.1] s= 0.0751 s
Similarly
2.5=3.1*sin [2*pi*f*t1] gives t1={1/(2*pi*f)}sin^-1[2.5/3.1]=0.139 s
So T = t1 - t= 0.139 - 0.0751 = 0.0639 s

(d) If the mass is oscillating with a maximum speed of 45 m/s the amplitude = 45/(2*pi*f)

= 45/[2*3.14*1.07]= 6.696 m

(e) Maximum kinetic energy = maximum potential energy = 0.5*k*(amplitude)^2
amplitude will be reduced by a factor of sq rt(2)

or

The max k.e = 1/2 m [ A ?] ^2 is a constant.
that is A? is a constant.

A1 ?1 = A2 ?2
?2/ ?1 = A1/A2
?2/ ?1 = ? k2/k1= A1/A2

Given k2 = 2k1
? k2/k1 = ?2 = A1/A2
Hence A2 = A1 / ?2

The factor is 1/sqrt(2) = 70.71%

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