SHMHard
Question
One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring constant k. A mass m hangs freely from the free end of the spring. The area of cross-section and the Young′s modulus of the wire are A and Y respectively. If the mass is slighly pulled down and released, it will oscillate with a time period T equal to
Options
A.2π(m/k)1/2
B.
C.2π[(mYA / kL)1/2]
D.2π[(mY / YA)1/2]
Solution
Keq =
= 
= 
∴ T =
= 
Note : Equivalent force constant for a wire is given by k =
. Because in case of a wire, F = 
ᐃL and in case of spring F = k. ᐃL. Comparing these two, we kind k of wire = 
Create a free account to view solution
View Solution FreeMore SHM Questions
How long after the beginning of motion is the displacement of a harmonically oscillating particle equal to one half its ...The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decreas...Length of a sonometer wire is 1.21 m. The three segments for fundamental frequency in the ratio 1 : 2 : 3....The P.E. of an oscillating particle at rest position is 15 J and its average K.E. is 5 J. The total energy of particle a...Masses m and 3 m are attached to the two ends of a spring of spring constant k. If the system vibrates freely, the perio...