Answer:
Maximum acceleration will be
Explanation:
We have given mass of the object m = 2 kg
Spring constant k = 55.6 N/m
Amplitude is given as A = 0.045 m
We know that maximum acceleration in SHM is given by
Maximum acceleration
We know that
So maximum acceleration =
Explanation:
Given that,
Distance 1, r = 100 m
Intensity,
If distance 2, r' = 25 m
We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :
.........(1)
Let I' is the intensity at r'. So,
............(2)
From equation (1) and (2) :
Intensity level is given by :
,
dB = 32.96 dB
Hence, this is the required solution.
Answer:
Explanation:
Given
Lowest four resonance frequencies are given with magnitude
50,100,150 and 200 Hz
The frequency of vibrating string is given by
where n=1,2,3 or ...n
L=Length of string
T=Tension
Mass per unit length
When string is clamped at mid-point
Effecting length becomes
Thus new Frequency becomes
i.e. New frequency is double of old
so new lowest four resonant frequencies are 100,200,300 and 400 Hz