6 Lattice and the Continuum
53
Fig. 6.2 Transversal deflection q of the masses of a linear chain in dependence on its position on
the x-axis. Because the horizontal forces cancel each other, according to our approximation (69),
the interaction forces F ki can be seen as purely transversal (see explanations in the text)
Here, we make a principle assumption of approximation. An angle α is defined
according to tan α = q i //x, with the difference q i of the deflections of two
neighbouring particles, q i = q i+1 − q i and their distance x = x i+1 − x i . We can
assume of all of these angles α that they are ‘small’; in other words„
α ≈ sin α ≈ tan α ≈
q
x
,
cos α ≈ 1 .
(69)
Now, we can immediately see out of Fig. 6.2 that for all tangential forces F x =
σ cos α is valid. Because the forces acting on every particle of its direct neighbours
have opposite directions, the tangential forces cancel each other due to the approximation (69). We therefore do not need to observe these for a motion. Therefore, all
the forces in Eq. (57), where we dropped the external forces F a , are purely transversal and lead to pure transversal motions. In case (b), for the transversal oscillations,
we read in Eqs. (63) all the deflections q i and all the forces F ki as pure transversal
quantities. With this additional explanation, we can literally practically use all the
details of the case (a) for longitudinal oscillations.
The single transversal force F i+1 i that acts on a single mass x from the right is per
definition equal to the (transversal) tension τ (x i , t), and for the sum of both left and
right transversal forces acting on a mass x i , the following is once again valid,
F i−1 i + F i+1 i = −F i i−1 + F i+1 i = −τ (x i−1 , t) + τ (x i , t) ,
with the result (64) when x is sufficiently small. We also arrive at Eqs. (65)–(67)
in exactly the same way as shown above. Hooke’s law (68) can be directly seen in
Fig. 6.2. For the transversal force, for example on a particle at x i , the following is
valid:
53
Fig. 6.2 Transversal deflection q of the masses of a linear chain in dependence on its position on
the x-axis. Because the horizontal forces cancel each other, according to our approximation (69),
the interaction forces F ki can be seen as purely transversal (see explanations in the text)
Here, we make a principle assumption of approximation. An angle α is defined
according to tan α = q i //x, with the difference q i of the deflections of two
neighbouring particles, q i = q i+1 − q i and their distance x = x i+1 − x i . We can
assume of all of these angles α that they are ‘small’; in other words„
α ≈ sin α ≈ tan α ≈
q
x
,
cos α ≈ 1 .
(69)
Now, we can immediately see out of Fig. 6.2 that for all tangential forces F x =
σ cos α is valid. Because the forces acting on every particle of its direct neighbours
have opposite directions, the tangential forces cancel each other due to the approximation (69). We therefore do not need to observe these for a motion. Therefore, all
the forces in Eq. (57), where we dropped the external forces F a , are purely transversal and lead to pure transversal motions. In case (b), for the transversal oscillations,
we read in Eqs. (63) all the deflections q i and all the forces F ki as pure transversal
quantities. With this additional explanation, we can literally practically use all the
details of the case (a) for longitudinal oscillations.
The single transversal force F i+1 i that acts on a single mass x from the right is per
definition equal to the (transversal) tension τ (x i , t), and for the sum of both left and
right transversal forces acting on a mass x i , the following is once again valid,
F i−1 i + F i+1 i = −F i i−1 + F i+1 i = −τ (x i−1 , t) + τ (x i , t) ,
with the result (64) when x is sufficiently small. We also arrive at Eqs. (65)–(67)
in exactly the same way as shown above. Hooke’s law (68) can be directly seen in
Fig. 6.2. For the transversal force, for example on a particle at x i , the following is
valid: