First slide

Applications of SHM

Question

A mass m oscillates with simple harmonic motion with frequency $f = \frac{ω}{2 π}$ and amplitude A on a spring with constant K , therefore

Moderate

A

The total energy of the system is $\frac{1}{2} {KA}^{2}$
B

The frequency is $\frac{1}{2 π} \sqrt{\frac{K}{M}}$
C

The maximum velocity occurs, when x = 0
D

All the above are correct

Solution

Total energy is given by $\frac{1}{2} {KA}^{2}$ .
time period and frequency are $T = 2 π \sqrt{\frac{M}{K}} a n d f = \frac{1}{2 π} \sqrt{\frac{K}{M}}$
Velocity is max at mean position (x=0) and min (x=A)

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