The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant

The law can be expressed mathematically as follows;

$f = \dfrac{1}{2\cdot l} \cdot \sqrt{\dfrac{T}{m} }$

The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously

Therefore, the law is verified indirectly, by rearranging the above equation as follows;

$m = \dfrac{1}{ l^2} \cdot \dfrac{T}{4\cdot f^2} }$

From which it can be shown that the following relation holds with the limits of error in the experiment

m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²

Explanation:

8 0

3 years ago

John(body mass=160pounds) is taking off for a long jump. The average ground reaction force Fg at takeoff is 1400 N pointing forw

slega [8]

Answer:

The free body diagram of John is shown in the attached figure (in the FBD john's mass is supposed to be concentrated at his center of mass and FBD is made of center of mass)

b) As shown in the FBD the ground reaction forces are:

i) In X direction $F_{x}=1400cos(35^{o})=1146.81N$

ii) In Y direction $F_{y}=1400sin(35^{o})=803.0N$

c) The respective accelerations in x and y direction's is calculated by newton's second law as indicated under

$\sum F_{x}=ma_{x}\\\\\therefore a_{x}=\frac{\sum F_{x}}{m}=\frac{1146.8N}{72.57kg}=15.80m/s^{2}\\\\\sum F_{y}=ma_{y}\\\\\therefore a_{y}=\frac{\sum F_{y}}{m}=\frac{803.00-72.57\times 9.81}{72.57}=1.255m/s^{2}$

4 0

3 years ago