0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# A spring block system is put into SHM in two experiments. In the first experiment, the block is pulled from the equilibrium position through a distance d1 and then released. In the second experiment, it is pulled from the equilibrium position through a greater distance d2 and then released. For the given scenario choose the incorrect statement.

A
Time period is same for both SHM
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Frequency is same for both SHM
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Maximum kinetic energy is same for both SHM
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Angular frequency is same for both SHM
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C Maximum kinetic energy is same for both SHMThe time period for a spring-block system performing simple harmonic motion is, T=2π√mk ⇒f=12π√km Angular frequency is given by, ω=√km Above parameters T, f, ω are independent of amplitude but Max. Kinetic energy depends on the amplitude of oscillation. K.Emax=12mω2A2 In the second case, the amplitude of oscillation of block will be higher (d2>d1), hence maximum kinetic energy will be greater in this case.

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program