Question
Question: A 0.5 kg body performs simple harmonic motion with a frequency of 2 Hz and an amplitude of 8 mm. Fin...
A 0.5 kg body performs simple harmonic motion with a frequency of 2 Hz and an amplitude of 8 mm. Find the maximum velocity of the body, its maximum acceleration and the maximum restoring force to which the body is subjected.
Solution
When a body is undergoing simple harmonic motion, we can say that the acceleration and its net force is proportional to the displacement and acts in the opposite direction of displacement.The motion is oscillatory, that is it repeats itself at a fixed frequency.
Formulas used:
Velocity=Vmax=Aω
Acceleration=Amax=Aω2
Restoring force Fmax=mAmax
A is the amplitude and ωis the angular velocity. m is the mass of the body.
Complete step by step answer:
We are given that, the amplitude is 8mm, frequency is 2Hz and the mass of the body is 0.5kg.
We know that the maximum velocity of the body in SHM is
Vmax=Aω⇒Vmax=A×(2πf)⇒Vmax=4π×0.008=0.101m/s
Hence the value of the velocity is 0.101m/s
We know that the maximum acceleration of the body in SHM is
Amax=Aω2⇒Amax=A×(2πf)2⇒Amax=4π×(0.008)2=1.264m/s2
Hence the value of the acceleration is 1.264m/s2
Maximum restoring force is
Fmax=mAmax∴Fmax=0.5×1.264=0.632N
Therefore, we have found all the required values that have been asked in our question.
Note: The applications of simple harmonic motion is widespread in various areas, especially in our day to day lives including clocks where the pendulum and the constant period oscillators work under SHM, shock absorbers in cars, musical instruments like violin or guitar, musical aiding instruments like metronome that act like a pendulum and sporting activities like bungee jumping and diving boards near pools.