Simple harmonic motion quiz 1

A mass 95 g is suspended from a spring with constant k=120.0 N/m. The mass is then lifted 1.75 cm from its equilibrium position and released. Determine the following.

Equations as functions of time for the position  $x\left(t\right)$  , velocity  $v\left(t\right)$  , and acceleration of the mass  $a\left(t\right)$  .

•  $v\left(t\right)=-0.6\sin\left(35.5t\right)$   $x\left(t\right)=0.015\sin\left(35.5t\right)$   $a\left(t\right)=-22.1\cos\left(35.5t\right)$  

•  $a\left(t\right)=-20.3\cos\left(30.1t\right)$   $x\left(t\right)=0.0175\sin\left(30.1t\right)$   $v\left(t\right)=-30.1\cos\left(30.1t\right)$  

•  $x\left(t\right)=-0.0175\sin\left(32.5t\right)$   $v\left(t\right)=22.5\cos\left(32.5t\right)$   $a\left(t\right)=-22.05\cos\left(32.5t\right)$  

•  $x\left(t\right)=1.75\cos\left(35.5t\right)$   $v\left(t\right)=-20.4\sin\left(35.5t\right)$   $a\left(t\right)=-2.5\cos\left(35.5t\right)$  

A certain mass is pulled down 7.7 cm and released. The mass then returns to its lowest point 18 more times in the first 100 s after its release. Determine the amplitude  $x_0$  , frequency  $f$  , and period of motion  $T$  .

• It is not possible to find those values with the given data.
  

•  $x_0=0.77\ m$   $f=5.56\ Hz$   $T=0.17\ s$  

•  $x_0=0.015\ m$   $f=6.55\ Hz$   $T=0.18\ s$  

•  $T=5.56\ s$   $x_0=0.077\ m$   $f=0.18\ Hz$  

If the length of a simple pendulum is twentyfivefold , how does the period change? Remember that the relationship between the period  $T$ , and the length  $L$  of a simple pendulum are related by the formula  $T^2=4\ \pi^{2\ }\frac{L}{g}$  

•  $25T$  

•  $5T$  

•  $T$  

• It can not be determined
  

In a certain oscillating system, the amplitude of the oscillation is 3.00 cm, and the maximum speed of the 200 g mass in the system is 15.0 cm/s.

Determine the spring constant and the maximum acceleration of the mass.

Remember that energy is conserved in this kind of system. Thus, you can use the formula  $\frac{1}{2}mv^2=\frac{1}{2}kx^2$  





•  $k=0.75\ \frac{N}{m}$   $a=5\ \frac{m}{s^2}$  

•  $k=0.6\ \frac{N}{m}$   $a=7.5\ \frac{m}{s^2}$  

•  $k=5\ \frac{N}{m}$   $a=0.75\ \frac{m}{s^2}$  

•  $k=6\ \frac{N}{m}$   $a=0.25\ \frac{m}{s^2}$  

If the mass of the bob on a pendulum is increased, how is the period affected?

• It is doubled

• It remains the same

• it can not be determined

• Why would you ask something like that?