Question

A physicist of 55 kg stands on a bathroom scale (a spring scale, with an internal spring).
She observes that when she mounts the scale suddenly, the pointer of the scale first
oscillates back and forth a few times with a frequency of 2.4 Hz. [5]
(a) What value of the spring constant can she deduce from these data?
(b) If she then takes a child of 20 kg in her arms and again stands on the scale, what
will be the new frequency of oscillation of the pointer?

Answer 1

a) Since the frequency is $\nu =\frac {1} {2\times\pi }\times\sqrt {\frac {k} {m}}$ , the physicist will be able to determine the elastic coefficient of the spring.

$k=4\times \pi^2 \times \nu^2 \times m= 4 \times \pi^2 \times 2.4^2 \times 55\approx 12506.76$ N/m.

b) $\nu=\frac{1}{2\times\pi}\times\sqrt{\frac{12506.76}{55+20}}\approx 2.06$ Hz